Question

A 200 g solid plastic ball (i.e., an insulator) with a 30 cm radius has a total charge of -20 μC that is uniformly distributed throughout its volume. The center of the ball is at the origin. A –5 μC point charge with a mass of 30 g is placed on the x-axis, 100 cm away from the center of the ball. Both charged objects are held in place.

a) Find the electric field (magnitude and direction) at the following points:

      i) at the origin                 ii) x=15 cm, y=0             iii) x=0, y=50 cm                  iv) x=150 cm, y=0      

b) Find the electric potential at the following points (taking the potential to be zero at large distances from the charges):

                      i) x=10 cm, y=0          ii) x=0, y=50 cm      

c) If the point charge is released, how fast will it be moving when it is a very large distance from the plastic ball?

Answer 1

As we can see that in this problem two charges are given having a -20µC on a plastic ball and -5µC charge as a point charge and we have to calculate electric field and potential at different points which I have shown below:

Part A

(i) in this case we have to find the E at orign so at origin there is no charge and then E=0

(ii) In this case we have to find at x=15cm,y=0;

As we know the charge is continuously distributed by using continuous charge distribution we can find the charge enclosed in the region of x=15cm and then we can find the electric field due to that reigion at x=15cm .

(iii) In this situation we have x=0cm, y= 50 cm

In this point both the charges will exert there electric field so have to calculate the net electric field at this point due to both charges .

As we can see that due to a plastic ball there is a large effect of electric field at this point because there is two factors which are dominating one is it has a small distance as compare to -5µC and other has it has a large charge value 4 times that of point charge all these factors show that the electric field at y= 50 cm is majorly due to a plastic ball but I have showed the electric field due to a point charge in this the electric field will resolve into two rectangular components and the EcosƟ and EsinƟ, the EsinƟ will add up in due to plastic ball but EcosƟ will be in +x direction so which is shown below.

(iv) When x=150 cm , y=0

This will be a linear charge case can be easily calculated as shown below

Part b

Potential at different points

Case (i)

When x = 10 cm

Then again using continuous charge distribution we can calculate the potential at x= 10cm

Case (ii)

When y = 50cm

Using the principle of superposition we can calculate the potential at y=50 cm

It’s a scalar quantity so we don’t have to resolve this in two components so can be easily shown below.

Part c

At large distances The potential is zero and r approach to infinity so as we can see that there ll be no force acting on the charge and then it ll not accelerate towards the plastic ball and it will remain at rest.

If you have any query regarding question please write in a comment. I hope you like it . Thank you.