Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Two identical conducting spheres (of negligible radius), having charges of opposite sign, attract each other with a force of

0.108 N

when separated by

0.5 meter

.The spheres are connected by a conducting wire, which is then removed (when the charge stops flowing), and thereafter repel each other with a force of

0.036 N

keeping the distance the same. What were the initial charges on the spheres?

Accepted Answer

Let initial charges on each particle is $q_{1} & q_{2}$ are separated by the distance, $r = 0.5 m$ , then the force of attraction,

$F = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}} = 0.108 N . . (1) \Rightarrow q_{1} q_{2} = \frac{0.108 \times {(0.5)}^{2}}{9 \times 10^{9}} = 3 \times 10^{- 12}$

When they are connected with conducting wire then due to conduction the charge on both the spheres becomes, $\frac{q_{1} + q_{2}}{2}$ , then the new force, $F' = \frac{1}{4 π ε_{0}} \frac{{(\frac{q_{1} + q_{2}}{2})}^{2}}{r^{2}} = 0.036 N . . . (2)$ ,

then from the equation $1 & 2$ ,

$\frac{\frac{1}{4 π ε_{0}} \frac{{(\frac{q_{1} + q_{2}}{2})}^{2}}{r^{2}} =}{\frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r^{2}}} \frac{0.036 N}{0.108 N} \Rightarrow \frac{{(\frac{q_{1} + q_{2}}{2})}^{2}}{\frac{0.108 \times {(0.5)}^{2}}{9 \times 10^{9}}} = 0.33 \Rightarrow {(q_{1} + q_{2})}^{2} = \frac{0.33 \times 0.108 \times {(0.5)}^{2} \times 4}{9 \times 10^{9}} \Rightarrow (q_{1} + q_{2}) = \sqrt{\frac{0.33 \times 0.108 \times {(0.5)}^{2} \times 4}{9 \times 10^{9}}} = \sqrt{3.96 \times 10^{- 12}} \approx 2 \times 10^{- 6} C$

It means,

$q_{1} = \frac{3 \times 10^{- 12}}{q_{2}}$ ,

$\Rightarrow (q_{1} + \frac{3 \times 10^{- 12}}{q_{1}}) = 2 \times 10^{- 6} C \Rightarrow {(q_{1})}^{2} - q_{1} \times 2 \times 10^{- 6} + 3 \times 10^{- 12} = 0 \Rightarrow q_{1} = \pm 1 \times 10^{- 6} C & q_{2} = \pm 3 \times 10^{- 6} C$

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