Question

17. Charge q1 = + 6.0 n C is on y-axis at y = + 3 cm
and charge q2 = - 6.0 n C is on y-axis at y = - 3cm. Calculate force on a test charge q0 = 2 nC
placed on X-axis at x = 4 cm.​

Answer 1

Given-Charge $q1 = + 6.0 n$ C is on y-axis at $y = + 3 cm$.

charge $q2 = - 6.0 n$ C is on y-axis at $y = - 3cm.$

Find-Calculate force on a test

$3^2+^2=5^2=R^2$

$Sin\theta=\frac{3}{5}$

$E_{net}=2Esin\theta=\frac{2\times kq_1q_2}{R^2}\times \frac{3}{5}$

$=2\times \frac{3}{5}\times \frac{9\times 10^9\times 6\times ^{-9}\times 3\times 10^{-9}}{(5\times 10^{-2})^2}$

$=5.184\times 10^{-5}$

= $51.84\muN$ (in negative -y direction)

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