Electric Charges and Fields NEET PYQ with Solutions

Q 1 :
Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as: [2025]

$\frac{F}{2}$
$\frac{3 F}{8}$
$\frac{3 F}{5}$
$\frac{2 F}{3}$

1

2

3

4

(2)

According to Coulomb's law, $F = \frac{k q^{2}}{R^{2}}$

By touching an uncharged sphere C with charged sphere A, charge on each sphere A and C becomes $\frac{q}{2}$ .

Now, when charged sphere C is kept in contact with sphere B,
then total charge is, $\frac{q}{2} + \frac{q}{1} = \frac{3 q}{2}$

Then,

Charge on B = $\frac{3 q}{4}$

Charge on C = $\frac{3 q}{4}$

Now, charge on A is $\frac{q}{2}$ and charge on B is $\frac{3 q}{4}$ .

So, $F' = \frac{k}{R^{2}} (\frac{q}{2}) (\frac{3 q}{4}) = k \frac{3}{8} \frac{q^{2}}{R^{2}} or F' = \frac{3}{8} F$

Q 2 :
Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes [2019]

$\frac{4 F}{3}$
$F$
$\frac{9 F}{16}$
$\frac{16 F}{9}$

1

2

3

4

(3)

In Case I :

$F = - \frac{1}{4 π ε_{0}} \frac{Q^{2}}{r^{2}}$ ...(i)

In Case II : $Q_{A} = Q - \frac{Q}{4}, Q_{B} = - Q + \frac{Q}{4}$

$∴$ $F' = \frac{1}{4 π ε_{0}} \frac{(Q - \frac{Q}{4}) (- Q + \frac{Q}{4})}{r^{2}}$

$= \frac{1}{4 π ε_{0}} \frac{(\frac{3}{4} Q) (\frac{- 3}{4} Q)}{r^{2}} = - \frac{1}{4 π ε_{0}} \frac{9}{16} \frac{Q^{2}}{r^{2}}$ ...(ii)

From equations (i) and (ii), $F' = \frac{9}{16} F$

Q 3 :
Suppose the charge of a proton and an electron differ slightly. One of them is $- e$ , the other is $(e + Δ e)$ . If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance $d$ (much greater than atomic size) apart is zero, then $Δ e$ is of the order of

[Given: mass of hydrogen $m_{h} = 1.67 \times 10^{- 27}$ $kg$ ] [2017]

$10^{- 23}$ C
$10^{- 37}$ C
$10^{- 47}$ C
$10^{- 20}$ C

1

2

3

4

(2)

A hydrogen atom consists of an electron and a proton.

$∴$ $Charge on one hydrogen atom = q_{e} + q_{p} = - e + (e + Δ e) = Δ e$

Since a hydrogen atom carries a net charge $Δ e$

$∴$ $Electrostatic force, F_{e} = \frac{1}{4 π ε_{0}} \frac{{(Δ e)}^{2}}{d^{2}}$ ...(i)

will act between two hydrogen atoms.

The gravitational force between two hydrogen atoms is given as $F_{g} = \frac{G m_{h} m_{h}}{d^{2}}$ ...(ii)

Since the net force on the system is zero, $F_{e} = F_{g}$
Using eqns. (i) and (ii), we get

$\frac{{(Δ e)}^{2}}{4 π ε_{0} d^{2}} = \frac{G m_{h}^{2}}{d^{2}} or, {(Δ e)}^{2} = 4 π ε_{0} G m_{h}^{2}$

$= 6.67 \times 10^{- 11} \times {(1.67 \times 10^{- 27})}^{2} [\frac{1}{(9 \times 10^{9})}]$

$Δ e \approx 10^{- 37}$ C

Q 4 :
Two identical charged spheres suspended from a common point by two massless strings of lengths $l$ , are initially at a distance $d (d < < l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v$ . Then $v$ varies as a function of the distance $x$ between the spheres, as [2016]

$v \propto x^{- 1 / 2}$
$v \propto x^{- 1}$
$v \propto x^{1 / 2}$
$v \propto x$

1

2

3

4

(1)

From figure, $T \cos θ = m g$ ...(i)

$T \sin θ = \frac{k q^{2}}{x^{2}}$ ...(ii)

From eqns. (i) and (ii), $\tan θ = \frac{k q^{2}}{x^{2} m g}$

Since $θ$ is small, $∴$ $\tan θ \approx \sin θ = \frac{x}{2 l}$

$∴$ $\frac{x}{2 l} = \frac{k q^{2}}{x^{2} m g} \Rightarrow q^{2} = x^{3} \frac{m g}{2 l k} or q \propto x^{3 / 2}$

$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} \frac{d x}{d t} = \frac{3}{2} \sqrt{x} v$

Since, $\frac{d q}{d t} = constant$ ; $∴$ $v \propto \frac{1}{\sqrt{x}}$

Topic Question Set

Q 2 :
Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes [2019]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 2 : Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes [2019]

Want Us to Email you About Special Offers & Updates?

Q 2 :
Two point charges A and B, having charges +Q and –Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes [2019]