Question

A very small ball has a mass of 5.00 x 10-3 kg and a charge of 4.00 µC. What magnitude electric field directed upward will balance the weight of the ball so that the ball is suspended motionless above the ground? a. 8.21 x 102 N/C b. 1.22 x 104 N/C c. 2.00 x 10-2 N/C d. 5.11 x 106 N/C

Answer 1

Answer:

• Option (b) 1.22 × 10⁴ N/C

By the magnitude of 1.25 × 20⁴ N/C electric field directed upward will balance the weight of the ball so that the ball is suspended motionless above the ground.

Explanation:

We know that,

$\implies \sf qE = mg$

$\implies \sf E = \dfrac{mg}{q}$

$\implies \sf E = \dfrac{5 \times 10^{-3} \times 10}{4 \times 10^{-6}}$

$\implies \sf E = \dfrac{5 \times 10^3 \times 10}{4}$

$\implies \sf E = 1.25 \times 10^4 \: N/C$

$\therefore$ By the magnitude of 1.25 × 20⁴ N/C electric field directed upward will balance the weight of the ball so that the ball is suspended motionless above the ground.