Question

The tiny ball at the end of thread shown in figure has a mass of 0.5 gram and is placed in a horizontal electric field of intensity 500 N/C. It is in equilibrium in position shown. The magnitude and the sign on charge of the ball is

Answer 1

The magnitude of charge is 17 micro C and the sign on charge of the ball is positive

Given : Mass of the tiny ball is 0.5 g

Magnitude of electric field is 500 N/C

To Find : The magnitude and the sign on charge of the ball

Solution : The magnitude of charge is 17 mu C and the sign on charge of the ball is positive

It is given that a tiny ball at the end of thread shown in figure has a mass of 0.5 gram and is placed in a horizontal electric field of intensity 500 N/C. It is in equilibrium in position shown.

We have to find the magnitude and sign on charge of the ball

In the diagram shown below

the tension in the ball has two component T cos(theta) (vertical) and T sin(theta) (horizontal)

now

m = 0.5 g

E = 500 N/C

theta = 30 °

q = ?

Tcos(30°) = mg    1)

Tsin(30°) = q E     2)

Dividing 1 and 2

tan(30) = $\frac{qE}{mg}$

$\sqrt{3}$ × $\frac{mg}{E}$ = q

q = $\frac{\sqrt{3} (10^{-3}) 10 }{1000}$

q = $\sqrt{3}$ × $10^{-5}$

q = 1.7 × $10^{-5}$

q = + 1.7 micro C

Here + sign is because the direction of electric field and force is same

So the magnitude of charge is 17 mu C and the sign on charge of the ball is positive

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