Physics question-finding the magnitude of the electric field

[nba23]

Physics question--finding the magnitude of the electric field

Homework Statement

A charged insulating ball of mass 7.0g with a uniform charge of 1.5mC hangs from a light thread inclined at an angle of 8 degrees to the vertical. Calculate the magnitude of the electric field.

Homework Equations

q=mg/e

e= Fe/q

The Attempt at a Solution

I tried to use e=mg/q and i kept getting the wrong answer. I am not sure how or if i should use the given angle.Im confused because i don't know which formula i need to use for this question.

 

[Doc Al]

I tried to use e=mg/q and i kept getting the wrong answer.

That would apply if the ball were suspended solely by the electric field, but its not--there's a string attached.

I am not sure how or if i should use the given angle.Im confused because i don't know which formula i need to use for this question.

Which way does the field point? (I presume horizontally.)

What forces act on the ball? (Draw a diagram for yourself.) What must the net force be? Hint: Consider vertical and horizontal force components separately.

 

[thomate1]

Dear student,

In order to find the magnitude of the electric field in this scenario, we need to use the formula e= Fe/q, where e is the magnitude of the electric field, Fe is the electric force, and q is the charge of the insulating ball.

First, we need to find the electric force acting on the ball. This can be done using the formula Fe= qE, where E is the electric field. In this case, the electric force is equal to the weight of the ball, which is given by mg. Therefore, we can write:

Fe= qE = mg

Solving for E, we get:

E= Fe/q = mg/q

Next, we need to find the value of q. We are given that the ball has a charge of 1.5mC, which is equivalent to 1.5x10^-3 C.

Plugging in the values, we get:

E= (0.007kg)(9.8m/s^2)/(1.5x10^-3 C) = 45.33x10^3 N/C

Therefore, the magnitude of the electric field is 45.33x10^3 N/C.

As for the given angle, we do not need to use it in this calculation as it does not affect the magnitude of the electric field. It only affects the direction of the electric field.

I hope this helps clarify the solution for you. Keep up the good work in your studies!