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Electrostatics Part 1

Section: Physics

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Question : 86 of 100

Marks: +1, -0

Two spherical conductors

A

B

1\ \text{mm}

2\ \text{mm}

are separated by a distance of

5\ \text{cm}

and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres

A

B

$4:1$
$1:2$
$2:1$
$1:4$

Solution:

After connection,

V_1=V_2

\; \Rightarrow K \; \frac{Q_1}{r_1}=K \; \frac{Q_2}{r^2}

\; \Rightarrow \; \frac{Q_1}{r_1}=\; \frac{Q_2}{r_2}

The ratio of electric fields

\;\; \frac{E_1}{E_2}=\; \frac{K \frac{Q_1}{r_1^{2}}}{K \; \frac{Q_2}{r_2^{2}}}=\; \frac{Q_1}{r_1^{2}} \times \; \frac{r_2^{2}}{Q_2}

\; \Rightarrow \; \frac{E_1}{E_2}=\; \frac{r_1 \times r_2^{2}}{r_1^{2} \times r_2} \Rightarrow \; \frac{E_1}{E_2}=\; \frac{r_2}{r_1}=\; \frac{2}{1}

Since the distance between the spheres is large as com effects may be ignored.

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