Question

A charged cork ball is suspended on a light string in the presence of a uniform electric field E~ = (2.2 × 105 N/C)ˆı + (3.6 × 105 N/C) ˆ . The ball is in equilibrium in the field. Find the charge on the ball. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of nC.

Answer 1

Complete Question:

A charged cork ball is suspended on a light string in the presence of a uniform electric field E~ = (2.2 × 105 N/C) i + (3.6 × 105 N/C) j . The ball is in equilibrium in the field. Find the charge on the ball. The acceleration due to gravity is 9.8 m/s 2. The mass of the ball is 0.7g. The length of the string is 1.4m and at an angle of 37 degrees from the center. Answer in units of nC.

Answer:

The charge on the ball is 0.1052 nC

Explanation:

Given;

vertical component of the Electric field, Ey = (3.6 × 10⁵ N/C)

horizontal component of the Electric field, Ex =  (2.2 × 10⁵ N/C)

If the cork ball is in equilibrium, then the upward force is equal to the downward force.

Also sum of all the forces acting on the cork ball is zero.

$Tsin \theta = q*E_x -------equation (i)\\\\Tcos \theta+q*E_y-mg= 0-----equation (ii)$

From equation (i)

Tsin(37) = (2.2 × 10⁵)q

$T = \frac{2.2X10^5*q}{sin(37)}--------equation (iii)$

Substitute equation (iii) into equation (ii) and solve for q

$\frac{2.2X10^5*q}{sin(37)}(cos (37)) + q((3.6X 10^5) -(7 X10^{-4} X 9.8) =0\\\\(291949.86)q +(360000)q = 0.00686\\\\(651949.86)q = 0.00686\\\\q = \frac{0.00686}{651949.86} = 1.0522 X 10^{-8} C = 0.1052nC$

Therefore, the charge on the ball is 0.1052 nC