## A positive charge is moved from a low potential point A to a high potential point B. Then the electric potential energy

Q: A positive charge is moved from a low potential point A to a high potential point B. Then the electric potential energy:

(a) Increases

(b) decreases

(c) will remain the same

(d) nothing definite can be predicted

Ans: (a)

Sol: $\large \Delta U = q (V_B – V_A )$

$\large \Delta U = q \Delta V$

As ΔV increases , so ΔU increases .

## Three charges 2q,-q,-q are located at the vertices of an equilateral triangle. At the centre of the triangle

Q: Three charges 2q , -q ,-q are located at the vertices of an equilateral triangle. At the centre of the triangle:

(a) The field is zero but potential is non – zero

(b) The field is non – zero but potential is zero

(c) Both field and potential are zero

(d) Both field and potential are non – zero

Ans: (b)

Sol: Let ABC is an equilateral triangle having centre O . Where OA = OB = OC = x (say)

Let three charges 2q , -q ,-q are located at the vertices A , B , C respectively .

Electric field at O due to 2q is $E_{OA} = \frac{1}{4 \pi \epsilon_0} \frac{2q}{x^2}$ (along AO produced)

Electric field at O due to B(-q ) is $E_{OB} = \frac{1}{4 \pi \epsilon_0} \frac{q}{x^2}$ (along OB)

and , Electric field at O due to C(-q) is $E_{OB} = \frac{1}{4 \pi \epsilon_0} \frac{q}{x^2}$ (along OC)

Now , find the resultant field these three field we can see that Electric field at O is not zero .

But , Electric Potential at O is ,

$\large V = \frac{1}{4 \pi \epsilon_0}\frac{2q}{x} + \frac{1}{4 \pi \epsilon_0}\frac{-q}{x} + \frac{1}{4 \pi \epsilon_0}\frac{-q}{x}$

V = 0

## Two equal -ve charges -q are fixed at the point (0,a) and (0,-a) on the y – axis. A positive charge Q is released from rest at the point (2a ,0) on the x- axis. The charge will

Q: Two equal -ve charges -q are fixed at the point (0,a) and (0,-a) on the y – axis. A positive charge Q is released from rest at the point (2a ,0) on the x- axis. The charge will

(a) Execute SHM about the origin

(b) Move to the origin and remain at rest

(c) Move to infinity

(d) Execute oscillatory but not SHM

Ans: (d)

## If two like charges of magnitude 1 × 10-9 coulomb and 9 × 10-9 coulomb are separated by a distance of 1 metre, then the point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is

Q: If two like charges of magnitude 1 × 10-9 coulomb and 9 × 10-9 coulomb are separated by a distance of 1 metre, then the point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is:

(a) 0.25 m from the charge 1 × 10-9 coulomb

(b) 0.75 m from the charge 9 × 10-9 coulomb

(c)Both 1 and 2

(d) At all points on the line joining the charges

Ans: (c)

Sol: Let AB be the line joining the charges A (1 × 10-9 C)  and B (9 × 10-9 C) . Let a charge q be placed at Point C  & force acting on it is zero and AC = x  , therefore BC = (1-x)

$F_{CA} = F_{CB}$

$\large \frac{1}{4 \pi \epsilon_0} \frac{q \times 1 \times 10^{-9}}{x^2}= \frac{1}{4 \pi \epsilon_0} \frac{q \times 9 \times 10^{-9}}{(1-x)^2}$

$\large \frac{1}{x^2} = \frac{9}{(1-x)^2}$

$\large \frac{1}{x} = \frac{3}{1-x}$

3x = 1-x ⇒ 4x = 1

x = 0.25 m

## A proton has a mass of 1.67 × 10-27 kg and charge 1.6 × 10-19 coulomb. If the proton is to be accelerated through a potential difference of one million volt, then the KE is

Q:    A proton has a mass of 1.67 × 10-27 kg and charge 1.6 × 10-19 coulomb. If the proton is to be accelerated through a potential difference of one million volt, then the KE is

(a) 1.6 × 10-15 J

(b) 1.6 × 10-13 J

(c) 1.6 × 10-25 J

(d) 3.2 × 10-13 J

Ans: (b)

Sol: $\large KE = \frac{1}{2} m v^2 = e V$

$\large KE = 1.6 \times 10^{-19} \times 10^6$

= 1.6 × 10-13 J

## Two concentric thin metallic spheres of radii R1 and R2 (R1 >R2 ) bear charges Q1 and Q2 respectively. Then the potential at r between R1 and R2 will be 1/(4πε0 ) times

Q: Two concentric thin metallic spheres of radii R1 and R2 (R1 >R2 ) bear charges Q1 and Q2 respectively. Then the potential at r between R1 and R2 will be 1/(4πε0 ) times

(a) $\large \frac{Q_1 + Q_2}{r}$

(b) $\large \frac{Q_1}{R_1}+ \frac{Q_2}{r}$

(c) $\large \frac{Q_1}{R_1}+ \frac{Q_2}{R_2}$

(d) $\large \frac{Q_2}{R_1}+ \frac{Q_1}{R_2}$

Ans: (b)

Sol: Let P be the point & O be the centre of concentric spheres , OP = r (say)

Potential at P  = Potential at P due to Q1 + Potential at P due to Q2

$\large V = \frac{1}{4\pi \epsilon_0} \frac{Q_1}{R_1} + \frac{1}{4\pi \epsilon_0} \frac{Q_2}{r}$

$\large V = \frac{1}{4\pi \epsilon_0} ( \frac{Q_1}{R_1} + \frac{Q_2}{r})$

## The force between two charges situated in air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is

Q: The force between two charges situated in air is F. The force between the same charges if the distance between them is reduced to half and they are situated in a medium having dielectric constant 4 is

(a) F/4

(b) 4F

(c) 16F

(d) F

Ans: (d)

Sol: $\large F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$  …..(i)

In medium ,

$\large F’ = \frac{1}{4\pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{(r/2)^2}$  …..(ii)

On dividing ,

$\large \frac{F’}{F} = \frac{4}{\epsilon_r} = \frac{4}{4}= 1$

F’ = F