Maths-
General
Easy

Question

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

  1. e to the power of x left parenthesis x plus 1 right parenthesis equals y
  2. straight e to the power of straight x left parenthesis straight x plus 1 right parenthesis plus 1 equals straight y
  3. e to the power of x left parenthesis x minus 1 right parenthesis plus 1 equals y
  4. None of these

The correct answer is: None of these

Book A Free Demo

+91

Grade*

Related Questions to study

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
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
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
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
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 1 y equals 1 is given by

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 1 y equals 1 is given by

Maths-General
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
General
Maths-

The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is :

The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is :

Maths-General