Calculate the work done to dissociate the system of the three charges placed on the vertices of a triangle as shown. (Here q = 1.6 x 10^–10 C)

Solution

Let, the charges on the vertices of triangle be q_A, q_B and q_C.
Given, charge q is 1.6 x 10^–10 C.
The work done to dissosciate the system would be the amount of potential energy required to bring a charge from one point to another.

$∴$ Potential energy is given by
$U = - (\frac{k q_{a} q_{b}}{AB} + k \frac{q_{b} q_{c}}{BC} + k \frac{q_{a} q_{c}}{AC}) and AB = BC = AC = 10 cm$

$\frac{1}{4 {πε}_{\circ}} [\frac{q . - 4 q}{0.10} + \frac{q . 2 q}{0.10} + \frac{(- 4 q) . (2 q)}{0.10}] = \frac{1}{4 π ε_{\circ}} [\frac{- 10 q^{2}}{0.10}] = - \frac{1}{4 π ε_{\circ}} (100 q^{2}) = - 9 \times 10^{9} \times 100 \times (1.6 \times 10^{- 10})^{2} = - 2.3 \times 10^{- 8} J$

Therefore, work done to dissosciate the system is W=-U
W= 2.3 $\times$ 10^-8 J

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Calculate the work done to dissociate the system of the three charges placed on the vertices of a triangle as shown. (Here q = 1.6 x 10^–10 C)