Three equal charges of $4.0 * 10^{-7}\;C$ are located at the corners of a right triangle whose sides are $6 \;cm$, $8 \; cm$ and $10 \; cm$ respectively. Find the force exerted on the charge located at $90^{\circ}$ angle.

Given, (need figure)

Three equal charge ($q_1 = q_2 = q_3 = 4 * 10^{-7}\;C)$

Distance $(AB) = 8\;cm = 0.08\;m$

Distance $(BC) = 6\;cm = 0.06 \;m$

Distance $(AC) = 10\;cm = 0.1\;m$

Angle $ፈ ABC = 90 ^{\circ}$

Force $(F) = ?$

$F_{AB} = $$\frac{1}{4 \pi \epsilon_0}\frac{q_1\;*\;q_2}{R^2}$$ = 9 * 10^9 \;*\; $$\frac{(0.4 \;*\; 10^{-6})^2}{(0.08)^2}$$ = 0.225\;N$

$F_{BC} = $$\frac{1}{4 \pi \epsilon_0}\frac{q_1\;*\;q_2}{R^2}$$ = 9 * 10^9 \;*\; $$\frac{(0.4 \;*\; 10^{-6})^2}{(0.06)^2}$$ = 0.40\;N$

Force exerted on the charge located at $90 ^{\circ}$ is,

$F = \sqrt{(F_{AB})^2 + (F_{BC})^2 + 2\;F_{AB}\;F_{BC}\;Cos\theta}$

$F = \sqrt{(0.225)^2 + (0.4)^2 + 2\;*\;0.225\;*\;0.4\;*\;Cos\;90}$

⇒ $F = 0.46\;N$

Return to Main menu