1. ⇒ (MHT CET 2023 13th May Morning Shift )

A charged spherical conductor of radius ' $R$ ' is connected momentarily to another uncharged spherical conductor of radius ' $r$ ' by means of a thin conducting wire, then the ratio of the surface charge density of the first to the second conductor is

A. $R : r^{2}$

B. $R : r$

C. $r : R$

D. $1 : 1$

Correct Option is (C)

$V_{1} = V_{2}$ ( $∵$ both the conductors have the same potential as they are connected)

$\begin{aligned} \Rightarrow \frac{q_{1}}{R} = \frac{q_{2}}{r} \\ σ = \frac{q}{4 π R^{2}} \end{aligned}$

$∴$ The ratio of their charge densities is:

$\begin{aligned} \Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{\frac{q_{1}}{R^{2}}}{\frac{q_{2}}{r^{2}}} \\ ∴ \frac{σ_{1}}{σ_{2}} = \frac{r}{R} \end{aligned}$

2. ⇒ (MHT CET 2023 13th May Morning Shift )

Three charges each of value $+ q$ are placed at the corners of an isosceles triangle $ABC$ of sides $AB$ and $AC$ each equal to $2 a$ . The mid points of $A B$ and $A C$ are $D$ and $E$ respectively. The work done in taking a charge $Q$ from $D$ to $E$ is ( $ε_{0} =$ permittivity of free space)

A. Zero

B. $\frac{3 q Q}{4 π ε_{0} a}$

C. $\frac{qQ}{8 π ε_{0} a}$

D. $\frac{3 qQ}{8 π ε_{0} a}$

Correct Option is (A)

$Given AB = AC = 2 a$

$V_{D} = V_{E}$ $(∵ D$ and $E$ are mid-points)

$∴$ The work done in taking a charge $q$ from $D$ to $E$ is

$W = q Δ V = q (V_{D} - V_{E})$

$∴ W = 0$ (Equipotential surfaces)

3. ⇒ (MHT CET 2023 12th May Evening Shift )

Two point charges ' $q 1$ ' and ' $q 2$ ' are separated by a distance ' $d$ '. What is the increase in potential energy of the system when ' $q 2$ ' is moved towards ' $q 1$ ' by a distance ' $x$ ' ? $(x < d) (\frac{1}{4 π ε_{0}} = K$ , constant)

A. $- \frac{K q_{1} q_{2} x}{d (d - x)}$

B. $- \frac{K q_{1} q_{2}}{d (d - x)}$

C. $\frac{{Kq}_{1} q_{2 x}}{(d^{2} - x^{2})}$

D. $\frac{{Kq}_{1} q_{2} x}{(d^{2} - x^{2})}$

Correct Option is (A)

The potential energy between two charges is given as $U = \frac{k q_{1} q_{2}}{r}$

Initial potential energy is $U_{f} = \frac{k q_{1} q_{2}}{r}$

When charge $q_{2}$ moves towards the $q_{1}$ the separation between the charges becomes $d - x$

The final potential energy is $U_{f} = \frac{{kq}_{1} q_{2}}{(d - x)}$

The increase in potential energy is

$\begin{aligned} ∴ Δ U = U_{f} - U_{f} \\ ∴ Δ U = \frac{k q_{1} q_{2}}{d} - \frac{k_{1} q_{2}}{(d - x)} \\ ∴ Δ U = {kq}_{1} q_{2} (\frac{1}{d} - \frac{1}{d - x}) \\ ∴ Δ U = \frac{- {kq}_{1} q_{2} x}{d (d - x)} \end{aligned}$

4. ⇒ (MHT CET 2023 12th May Evening Shift )

Three point charges $+ Q, + 2 q$ and $+ q$ are placed at the vertices of a right angled isosceles triangle. The net electrostatic potential energy of the configuration is zero, if $Q$ is equal to

MHT CET 2023 12th May Evening Shift Physics - Electrostatics Question 8 English

A. $- \frac{\sqrt{2}}{3} q$

B. $+ \frac{\sqrt{2}}{3} q$

C. $- \frac{3}{\sqrt{2}} q$

D. $+ \frac{3}{\sqrt{2}} q$

Correct Option is (A)

Net electrostatic potential energy of the system is,

$\begin{aligned} U = \frac{1}{4 π ε_{0}} (\frac{Q q}{a} + \frac{2 Q q}{a} + \frac{2 q q}{\sqrt{2} a}) = 0 \\ Q + 2 Q + \frac{2 q}{\sqrt{2}} = 0 \\ 3 Q + \sqrt{2} q = 0 \\ ∴ Q = \frac{- \sqrt{2} q}{3} \end{aligned}$

5. ⇒ (MHT CET 2023 11th May Morning Shift )

Two charges of equal magnitude ' $q$ ' are placed in air at a distance ' $2 r$ ' apart and third charge ' $- 2 q$ ' is placed at mid point. The potential energy of the system is $(ε_{0} =$ permittivity of free space)

A. $- \frac{q^{2}}{8 π ε_{0} r}$

B. $- \frac{3 q^{2}}{8 π ε_{0} r}$

C. $- \frac{5 q^{2}}{8 π ε_{0} r}$

D. $- \frac{7 q^{2}}{8 π ε_{0} r}$

Correct Option is (D)

Potential energy of ' $n$ ' point charges,

$U = \frac{1}{4 π ε_{0}} \sum_{all pairs} \frac{q_{j} q_{k}}{r_{jk}}$

For 3 point charges,

$\begin{aligned} ∴ U & = - \frac{q (2 q)}{4 π ε_{0} r} - \frac{q (2 q)}{4 π ε_{0} r} + \frac{q (q)}{4 π ε_{0} (2 r)} \\ U & = - \frac{2 q^{2}}{4 π ε_{0} r} - \frac{2 q^{2}}{4 π ε_{0} r} + \frac{q^{2}}{4 π ε_{0} (2 r)} \\ U & = - \frac{4 q^{2}}{8 π ε_{0} r} - \frac{4 q^{2}}{8 π ε_{0} r} + \frac{q^{2}}{8 π ε_{0} r} \\ U & = - \frac{7 q^{2}}{8 π ε_{0} r} \end{aligned}$

⇒Electrostatics