A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it up. The charged object is in a uniform electric field of 4500 N/C and has a weight of 0.05N. What must the charge of the object be?

A. 6.67e-6 C
B. 8.89e-6 C
C. 1.11e-5 C
D. 1.33 e-5 C

The charge of the object must be

Option C

Explanation:

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.