Physics-
General
Easy

Question

In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

  1. Negative, increase    
  2. Positive, decrease    
  3. Negative, decrease    
  4. Positive, increase    

The correct answer is: Positive, increase


    Since, the proton is moving against the direction of electric field so, work is done by the proton against electric field. It implies that electric field does negative work on the proton.
    Again, proton is moving in electric field from low potential region to high potential region hence, its potential energy increases.

    Related Questions to study

    General
    maths-

    Solution of fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent is

    Solution of fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent is

    maths-General
    General
    physics-

    Two charges q subscript 1 end subscriptand q subscript 2 end subscript are placed 30 cm apart, as shown in the figure. A third charge q subscript 3 end subscriptis moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma where k is

    When charge q subscript 3 end subscriptis at C, then its potential energy is
    U subscript C end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    Where charge q subscript 3 end subscriptis at D, then
    U subscript D end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction close parentheses
    Hence, change in potential energy
    increment U equals U subscript D end subscript minus U subscript C end subscript
    equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    b u t blank increment U equals fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k
    therefore fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    ⟹ k equals q subscript 2 end subscript open parentheses 10 minus 2 close parentheses equals 8 q subscript 2 end subscript

    Two charges q subscript 1 end subscriptand q subscript 2 end subscript are placed 30 cm apart, as shown in the figure. A third charge q subscript 3 end subscriptis moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma where k is

    physics-General
    When charge q subscript 3 end subscriptis at C, then its potential energy is
    U subscript C end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    Where charge q subscript 3 end subscriptis at D, then
    U subscript D end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction close parentheses
    Hence, change in potential energy
    increment U equals U subscript D end subscript minus U subscript C end subscript
    equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    b u t blank increment U equals fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k
    therefore fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    ⟹ k equals q subscript 2 end subscript open parentheses 10 minus 2 close parentheses equals 8 q subscript 2 end subscript
    General
    physics-

    Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential differenceV subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript. If t subscript 1 end subscriptand t subscript 2 end subscriptbe the distance between them. Then

    Potential difference between two equipotential surfaces A and B.
    V subscript A end subscript minus V subscript B end subscript equals k q open parentheses fraction numerator 1 over denominator r subscript A end subscript end fraction minus fraction numerator 1 over denominator r subscript B end subscript end fraction close parentheses
    equals k q open parentheses fraction numerator r subscript B end subscript minus r subscript A end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction close parentheses
    equals fraction numerator k q t subscript 1 end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction
    Or
    t subscript 1 end subscript equals fraction numerator open parentheses V subscript A end subscript minus V subscript B end subscript close parentheses r subscript A end subscript r subscript B end subscript over denominator k q end fraction
    Or t subscript 1 end subscript proportional to r subscript A end subscript r subscript B end subscript
    Similarly, t subscript 2 end subscript proportional to r subscript B end subscript r subscript C end subscript
    Since,
    r subscript A end subscript less than r subscript B end subscript less than r subscript C end subscript comma therefore r subscript A end subscript r subscript B end subscript less than r subscript B end subscript r subscript C end subscript
    therefore blank t subscript 1 end subscript less than t subscript 2 end subscript

    Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential differenceV subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript. If t subscript 1 end subscriptand t subscript 2 end subscriptbe the distance between them. Then

    physics-General
    Potential difference between two equipotential surfaces A and B.
    V subscript A end subscript minus V subscript B end subscript equals k q open parentheses fraction numerator 1 over denominator r subscript A end subscript end fraction minus fraction numerator 1 over denominator r subscript B end subscript end fraction close parentheses
    equals k q open parentheses fraction numerator r subscript B end subscript minus r subscript A end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction close parentheses
    equals fraction numerator k q t subscript 1 end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction
    Or
    t subscript 1 end subscript equals fraction numerator open parentheses V subscript A end subscript minus V subscript B end subscript close parentheses r subscript A end subscript r subscript B end subscript over denominator k q end fraction
    Or t subscript 1 end subscript proportional to r subscript A end subscript r subscript B end subscript
    Similarly, t subscript 2 end subscript proportional to r subscript B end subscript r subscript C end subscript
    Since,
    r subscript A end subscript less than r subscript B end subscript less than r subscript C end subscript comma therefore r subscript A end subscript r subscript B end subscript less than r subscript B end subscript r subscript C end subscript
    therefore blank t subscript 1 end subscript less than t subscript 2 end subscript
    parallel
    General
    physics-

    Charges +q and –q are placed at points A and B respectively which are a distance 2 Lapart, C is the mid-point between A and B. The work done in moving a charge+Q along the semicircle CRD is

    In Ist case, when charge plus Q is situated at C

    Electric potential energy of system
    U subscript 1 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator L end fraction
    In IInd case, when charge plus Qis moved from C to D.

    Electric potential energy of system in that case
    U subscript 2 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator 3 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses open parentheses Q close parentheses over denominator L end fraction
    therefore W o r k blank d o n e equals increment U equals U subscript 2 end subscript minus U subscript 1 end subscript
    equals open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator 3 L end fraction minus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    negative open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction. open square brackets fraction numerator 1 over denominator 3 L end fraction minus fraction numerator 1 over denominator L end fraction close square brackets
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses 1 minus 3 close parentheses over denominator 3 L end fraction
    equals fraction numerator negative 2 q Q over denominator 12 pi epsilon subscript 0 end subscript L end fraction equals negative fraction numerator q Q over denominator 6 pi epsilon subscript 0 end subscript L end fraction

    Charges +q and –q are placed at points A and B respectively which are a distance 2 Lapart, C is the mid-point between A and B. The work done in moving a charge+Q along the semicircle CRD is

    physics-General
    In Ist case, when charge plus Q is situated at C

    Electric potential energy of system
    U subscript 1 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator L end fraction
    In IInd case, when charge plus Qis moved from C to D.

    Electric potential energy of system in that case
    U subscript 2 end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator open parentheses q close parentheses open parentheses negative q close parentheses over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q Q over denominator 3 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses negative q close parentheses open parentheses Q close parentheses over denominator L end fraction
    therefore W o r k blank d o n e equals increment U equals U subscript 2 end subscript minus U subscript 1 end subscript
    equals open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator 3 L end fraction minus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    negative open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator 2 L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q Q over denominator L end fraction close parentheses
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction. open square brackets fraction numerator 1 over denominator 3 L end fraction minus fraction numerator 1 over denominator L end fraction close square brackets
    equals fraction numerator q Q over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator open parentheses 1 minus 3 close parentheses over denominator 3 L end fraction
    equals fraction numerator negative 2 q Q over denominator 12 pi epsilon subscript 0 end subscript L end fraction equals negative fraction numerator q Q over denominator 6 pi epsilon subscript 0 end subscript L end fraction
    General
    physics-

    Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure left parenthesis a less than b less than c right parenthesis. Their surface charge densities are sigma comma negative sigma a n d sigma respectively. Calculate the electric potential on the surface of shell A

    The electric potential on the surface of shell
    V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript a end subscript over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction plus fraction numerator q subscript b end subscript over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript c end subscript over denominator c end fraction
    Or V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi a to the power of 2 end exponent sigma over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi b to the power of 2 end exponent open parentheses negative sigma close parentheses over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi c to the power of 2 end exponent sigma over denominator c end fraction
    open parentheses because q equals 4 pi r to the power of 2 end exponent sigma close parentheses
    o r blank V subscript A end subscript equals fraction numerator sigma over denominator epsilon subscript 0 end subscript end fraction left parenthesis a minus b plus c right parenthesis

    Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure left parenthesis a less than b less than c right parenthesis. Their surface charge densities are sigma comma negative sigma a n d sigma respectively. Calculate the electric potential on the surface of shell A

    physics-General
    The electric potential on the surface of shell
    V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript a end subscript over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction plus fraction numerator q subscript b end subscript over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript c end subscript over denominator c end fraction
    Or V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi a to the power of 2 end exponent sigma over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi b to the power of 2 end exponent open parentheses negative sigma close parentheses over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi c to the power of 2 end exponent sigma over denominator c end fraction
    open parentheses because q equals 4 pi r to the power of 2 end exponent sigma close parentheses
    o r blank V subscript A end subscript equals fraction numerator sigma over denominator epsilon subscript 0 end subscript end fraction left parenthesis a minus b plus c right parenthesis
    General
    physics-

    Work required to set up the four charge configuration (as shown in the figure) is

    Work is required to set up the four charge configuration
    q subscript 1 end subscript equals plus q comma blank q subscript 2 end subscript equals negative q comma blank q subscript 3 end subscript equals plus q a n d q subscript 4 end subscript equals negative q

    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator A B end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator B C end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator C D end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator D A end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses plus q close parentheses over denominator A C end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses negative q close parentheses over denominator B D end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets negative fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus square root of 2 close square brackets equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus 1.414 close square brackets
    W equals negative 0.21 cross times fraction numerator q to the power of 2 end exponent over denominator epsilon subscript 0 end subscript a end fraction open parentheses a p p r o x. close parentheses

    Work required to set up the four charge configuration (as shown in the figure) is

    physics-General
    Work is required to set up the four charge configuration
    q subscript 1 end subscript equals plus q comma blank q subscript 2 end subscript equals negative q comma blank q subscript 3 end subscript equals plus q a n d q subscript 4 end subscript equals negative q

    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator A B end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator B C end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator C D end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator D A end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses plus q close parentheses over denominator A C end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses negative q close parentheses over denominator B D end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets negative fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus square root of 2 close square brackets equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus 1.414 close square brackets
    W equals negative 0.21 cross times fraction numerator q to the power of 2 end exponent over denominator epsilon subscript 0 end subscript a end fraction open parentheses a p p r o x. close parentheses
    parallel
    General
    physics-

    The points resembling equal potentials are

    The points S and R are inside the uniform electric field, so these will be at equal potential.

    The points resembling equal potentials are

    physics-General
    The points S and R are inside the uniform electric field, so these will be at equal potential.
    General
    physics-

    In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript then

    When a positive charge is moved from one point to another in an electric of magnetic field, then under the influence of the field force acts on the particle and an external agent will have to do work against this force. But in the given case the charge moves under influence of no field, hence it does not experience any force therefore, no work is done.
    W subscript A end subscript equals W subscript B end subscript equals W subscript C end subscript equals 0

    In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript then

    physics-General
    When a positive charge is moved from one point to another in an electric of magnetic field, then under the influence of the field force acts on the particle and an external agent will have to do work against this force. But in the given case the charge moves under influence of no field, hence it does not experience any force therefore, no work is done.
    W subscript A end subscript equals W subscript B end subscript equals W subscript C end subscript equals 0
    General
    physics-

    Three charges Q subscript 0 end subscript comma negative q and – q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q subscript 0 end subscriptis equal to

    Here total electrostatic potential energy is zero
    i e comma blank fraction numerator negative Q subscript 0 end subscript q over denominator l end fraction minus fraction numerator q Q subscript 0 end subscript over denominator l end fraction plus fraction numerator q to the power of 2 end exponent over denominator square root of 2 l end root end fraction equals 0
    On solving,
    Q subscript 0 end subscript equals fraction numerator q over denominator 2 square root of 2 end fraction equals fraction numerator 2 q over denominator square root of 32 end fraction

    Three charges Q subscript 0 end subscript comma negative q and – q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q subscript 0 end subscriptis equal to

    physics-General
    Here total electrostatic potential energy is zero
    i e comma blank fraction numerator negative Q subscript 0 end subscript q over denominator l end fraction minus fraction numerator q Q subscript 0 end subscript over denominator l end fraction plus fraction numerator q to the power of 2 end exponent over denominator square root of 2 l end root end fraction equals 0
    On solving,
    Q subscript 0 end subscript equals fraction numerator q over denominator 2 square root of 2 end fraction equals fraction numerator 2 q over denominator square root of 32 end fraction
    parallel
    General
    physics-

    A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. V subscript A end subscript comma blank V subscript B end subscript comma blank V subscript c end subscriptbe the potentials at points A, B and C respectively. Then

    At each point on the surface of a conducting sphere the potential is equal.
    So, V subscript A end subscript equals V subscript B end subscript equals V subscript C end subscript

    A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. V subscript A end subscript comma blank V subscript B end subscript comma blank V subscript c end subscriptbe the potentials at points A, B and C respectively. Then

    physics-General
    At each point on the surface of a conducting sphere the potential is equal.
    So, V subscript A end subscript equals V subscript B end subscript equals V subscript C end subscript
    General
    physics-

    The figure shows electric potential V as a function of x. Rank the four regions according to the magnitude of x-component of the electric field E within them, greatest first

    Electric field
    E equals negative fraction numerator d V over denominator d x end fraction
    For I region, V subscript 1 end subscript=constant

    therefore blank fraction numerator d V subscript 1 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 1 end subscript equals 0 blank
    For II region,
    V subscript 2 end subscript equals plus v e equals plus f open parentheses x close parentheses
    therefore E subscript 2 end subscript equals negative fraction numerator d V subscript 2 end subscript over denominator d x end fraction equals negative v e
    For III region.
    V subscript 3 end subscript=constant
    therefore blank fraction numerator d V subscript 3 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 3 end subscript equals 0 blank
    For IV region, V subscript 1 end subscript equals negative f open parentheses x close parentheses
    therefore blank E subscript 4 end subscript equals negative fraction numerator d V subscript 4 end subscript over denominator d x end fraction equals plus v e
    From these values, we have
    E subscript 2 end subscript greater than E subscript 4 end subscript greater than E subscript 1 end subscript equals E subscript 3 end subscript

    The figure shows electric potential V as a function of x. Rank the four regions according to the magnitude of x-component of the electric field E within them, greatest first

    physics-General
    Electric field
    E equals negative fraction numerator d V over denominator d x end fraction
    For I region, V subscript 1 end subscript=constant

    therefore blank fraction numerator d V subscript 1 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 1 end subscript equals 0 blank
    For II region,
    V subscript 2 end subscript equals plus v e equals plus f open parentheses x close parentheses
    therefore E subscript 2 end subscript equals negative fraction numerator d V subscript 2 end subscript over denominator d x end fraction equals negative v e
    For III region.
    V subscript 3 end subscript=constant
    therefore blank fraction numerator d V subscript 3 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 3 end subscript equals 0 blank
    For IV region, V subscript 1 end subscript equals negative f open parentheses x close parentheses
    therefore blank E subscript 4 end subscript equals negative fraction numerator d V subscript 4 end subscript over denominator d x end fraction equals plus v e
    From these values, we have
    E subscript 2 end subscript greater than E subscript 4 end subscript greater than E subscript 1 end subscript equals E subscript 3 end subscript
    General
    maths-

    The general solution of the equation fraction numerator d y over denominator d x end fraction equals 1 plus x y is

    The general solution of the equation fraction numerator d y over denominator d x end fraction equals 1 plus x y is

    maths-General
    parallel
    General
    Maths-

    Solution of 2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x comma x equals pi over 2 comma y equals 1is given by

    G i v e n comma space 2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x
rightwards double arrow 2 y sin space x fraction numerator d y over denominator d x end fraction plus y squared cos space x equals sin 2 x
rightwards double arrow 2 sin x fraction numerator d over denominator d x end fraction y squared minus y squared fraction numerator d over denominator d x end fraction sin x equals sin 2 x
rightwards double arrow y squared fraction numerator d over denominator d x end fraction sin x space equals sin 2 x
i n t e g r a t i n g space b o t h space s i d e s comma
integral sin x space y squared space d x space equals integral sin 2 x space d x
rightwards double arrow sin x space y squared equals fraction numerator negative cos 2 x over denominator 2 end fraction plus C
N o w comma space f o r space x equals straight pi over 2 comma space y equals 1
rightwards double arrow sin straight pi over 2.1 squared equals fraction numerator negative cos 2 begin display style straight pi over 2 end style over denominator 2 end fraction plus C
rightwards double arrow C equals 1 plus open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses
rightwards double arrow C equals 1 half
sin x. y squared equals fraction numerator negative cos 2 x over denominator 2 end fraction plus 1 half
rightwards double arrow y squared equals fraction numerator 1 minus cos 2 x over denominator 2 sin x end fraction

    Solution of 2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x comma x equals pi over 2 comma y equals 1is given by

    Maths-General
    G i v e n comma space 2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x
rightwards double arrow 2 y sin space x fraction numerator d y over denominator d x end fraction plus y squared cos space x equals sin 2 x
rightwards double arrow 2 sin x fraction numerator d over denominator d x end fraction y squared minus y squared fraction numerator d over denominator d x end fraction sin x equals sin 2 x
rightwards double arrow y squared fraction numerator d over denominator d x end fraction sin x space equals sin 2 x
i n t e g r a t i n g space b o t h space s i d e s comma
integral sin x space y squared space d x space equals integral sin 2 x space d x
rightwards double arrow sin x space y squared equals fraction numerator negative cos 2 x over denominator 2 end fraction plus C
N o w comma space f o r space x equals straight pi over 2 comma space y equals 1
rightwards double arrow sin straight pi over 2.1 squared equals fraction numerator negative cos 2 begin display style straight pi over 2 end style over denominator 2 end fraction plus C
rightwards double arrow C equals 1 plus open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses
rightwards double arrow C equals 1 half
sin x. y squared equals fraction numerator negative cos 2 x over denominator 2 end fraction plus 1 half
rightwards double arrow y squared equals fraction numerator 1 minus cos 2 x over denominator 2 sin x end fraction
    General
    Maths-

    Equation of the curve passing through (3, 9) which satisfies the differential equation fraction numerator d y over denominator d x end fraction equals x plus 1 over x squared is

    Equation of the curve passing through (3, 9) which satisfies the differential equation fraction numerator d y over denominator d x end fraction equals x plus 1 over x squared is

    Maths-General
    General
    Maths-

    If f left parenthesis x right parenthesis space equals space f apostrophe left parenthesis x right parenthesis and f(1) = 2, then f(3) =

    If f left parenthesis x right parenthesis space equals space f apostrophe left parenthesis x right parenthesis and f(1) = 2, then f(3) =

    Maths-General
    parallel

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