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Physics · 2025 · 3 marks
CBSE 2025 · Region 5 · Set 3 · Q26
(a) Two small solid metal balls A and B of radii R and 2 R having charge densities 2 σ and 3 σ respectively are kept far apart. Find the charge…
Two small solid metal balls A and B of radii R and $\displaystyle 2$ R having charge densities $\displaystyle 2 \sigma$ and $\displaystyle 3 \sigma$ respectively are kept far apart. Find the charge densities on A and B after they are connected by a conducting wire.
Two infinitely long straight wires ' $\displaystyle 1$ ' and ' $\displaystyle 2$ ' are placed d distance apart, parallel to each other, as shown in the figure. They are uniformly charged having charge densities $\displaystyle \lambda$ and $\displaystyle -\frac{\lambda}{2}$ respectively. Locate the position of the point from wire ' $\displaystyle 1$ ' at which the net electric field is zero and identify the region in which it lies.
Marking-scheme solution
For ball A:
$\displaystyle q_{1}=2 \sigma \times 4 \pi R^{2}=8 \pi R^{2} \sigma$
For ball B:
$\displaystyle q_{2}=3 \sigma \times 4 \pi(2 R)^{2}=48 \pi R^{2} \sigma$
Total charge $\displaystyle (Q)=q_{1}+q_{2}=56 \pi R^{2} \sigma$
When balls A and B are connected by a wire, their potentials will be equal. Let q be the charge on ball A and $\displaystyle (Q-q)$ the charge on ball B after connecting the wire:
$\displaystyle \frac{K q}{R}=\frac{K(Q-q)}{2 R}$
$\displaystyle 2 q=Q-q$
$\displaystyle q=\frac{Q}{3}=\frac{56 \pi R^{2} \sigma}{3}$
$\displaystyle Q-\frac{Q}{3}=\frac{112 \pi R^{2} \sigma}{3}$
$\displaystyle \sigma_{A}=\frac{\dfrac{56 \pi R^{2} \sigma}{3}}{4 \pi R^{2}}=\frac{14}{3} \sigma$
$\displaystyle \sigma_{B}=\frac{\dfrac{112 \pi R^{2} \sigma}{3}}{4 \pi(2 R)^{2}}=\frac{7}{3} \sigma$
Electric field due to wire $\displaystyle 1$ and wire $\displaystyle 2$ at point P:
$\displaystyle E_{1}=\frac{\lambda}{2 \pi \varepsilon_{0} x}$
$\displaystyle E_{2}=\frac{\lambda}{2 \pi \varepsilon_{0}(x+d)}$
At P, net electric field is zero:
$\displaystyle E_{1}=E_{2}$
$\displaystyle \frac{\lambda}{2 \pi \varepsilon_{0} x}=\frac{\lambda}{2 \times 2 \pi \varepsilon_{0}(x+d)}$
$\displaystyle x=-2 d$
The negative sign indicates that the point lies in region C.
At a distance 2d from wire $\displaystyle 1$ the electric field is zero.
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