Name: Electric Charges and Electric Fields: A Review
Uploaded: 2019-07-19
Duration: 6 min 41 s
Description: In this video, we are going to review the electric charges, the law of charge, conservation of charges and electric field in detail. Watch this to learn more.

Question

Two conducting spheres

A

and

B

of the same size are situated at some distance apart. Each has a charge of

+ q

Coulomb and they repel each other with a force of

10^{- 5} N

. A third conducting uncharged sphere

C

of same size is first touched with $A$ and then with $B$ and then placed exactly between

A

and

B

. Calculate the resultant force on

C

.

Accepted Answer

Given,

$q_{1} = q_{2} = + q \Rightarrow q_{A} = q_{B} = + q F = 10^{- 5} N = \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{r^{2}} . . . . . . . . . (i)$

Third sphere is $C$ Which is uncharged.

When $C$ , becomes in contact with $A$ , then charge on $A$ and $C$ $= q_{A}^{'} = q_{C} = \frac{q + 0}{2} = \frac{q}{2}$

and when C becomes in contact with $B,$ then charge on $B$ and $C$ ${=q}_{B}^{'} = q_{c}^{'} = \frac{\frac{q}{2} + \frac{q}{1}}{2} = \frac{3 q}{4}$

Here $q'_{A} = q / 2, q_{B}^{'} = \frac{3 q}{4}, q_{C}^{'} = \frac{3 q}{4}$

Now $F_{net}$ at $C$ $= F_{CA} - F_{CB} = \frac{1}{4 π ϵ_{0}} . \frac{\frac{q}{2} \times \frac{3 q}{4}}{{(\frac{r}{2})}^{2}} - \frac{1}{4 π ϵ_{0}} . \frac{\frac{3 q}{4} \times \frac{3 q}{4}}{{(\frac{r}{2})}^{2}}$

$= \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{r^{2}} [\frac{3 \times 4}{8} - \frac{9 \times 4}{16}] = 10^{- 5} [\frac{6 - 9}{16}]$

$= - \frac{3 \times 4}{16} \times 10^{- 5} N = - 0.75 \times 10^{- 5} N = - 7.5 \times 10^{- 5} N$

Hence, magnitude of net force on sphere $= 7.5 \times 10^{- 5} N$ and $- ve$ sign indicates that direction of $F_{net}$ is forwards $A$ (or opposite to $B$ )

Important Questions on Electric Charges and Fields

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