Three equal charges, each having a magnitude of 2.0 x 10^-6 C, are placed at the three corners of a right angled triangle of sides 3 cm, 4 cm and 5 cm. The force (in magnitude) on the charge at the right angled corner is

50 N

26 N

29 N

45.9 N

Solution

45.9 N

Consider the diagram

q_A = q_B = q_C = 2 × 10^-6 C

The sides of right angle triangle

AB = 3 cm

BC = 4 cm

AC = 5 cm

The force on A due to B is

F_A = $\frac{1}{4 {πε}_{o}} \frac{q_{A} q_{B}}{{(AB)}^{2}}$

= 9 × 10⁹ × $\frac{{(2 \times 10^{- 6})}^{2}}{{(0.03)}^{2}}$

F_A = 40 N ( along BA)

The force on B due to C is

F_C = $\frac{1}{4 {πε}_{o}} \frac{q_{C} q_{B}}{{(BC)}^{2}}$

= $\frac{1}{4 {πε}_{o}} \frac{(2 \times 10^{- 6})}{{(4)}^{2}}$

F_C = 22.5 N ( along BC )

The resultant force on the charge at B,

F = $\sqrt{{F_{A}}^{2} + {F_{C}}^{2}}$

= $\sqrt{{(40)}^{2} + {(22.5)}^{2}}$

F = 45.9 N

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