Question

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-4Q 9Q 10 (m) Two point charges of magnitude 90 and-4Q where Q-0 are placed 10 meters apart on a line as shown. The electric field created by these two charges is zero at t a point P (not shown Question 1 (2 points) The point P could lie to the RIGHT of the charge -40 O True False

Answer 1

See the point is that the Electric field is towards the charge when the charge is negative and it is away from the charge when the charge is positive. Now in order to get the zero electric field on the line we need to get the directions of electric field in opposite direction and their magnitudes must also be equal.

1) the direction of electric field due to -4Q charge is rightwards to the left of it and leftwards to the right of it, while direction of electric field of 9Q charge is rightwards to the right of it and left wards to the left of it.

Hence direction of electric field due to both the charges are in the same direction in between the charges, therefore there cant exist a point where electric field is zero.

2) as we know that electric field is inversely proportional to square of the distance, therefore magnitude of electric field will decrease as we move away from the charge. If we see on the left of the 9Q charge the directions of the electric field are opposite but this time the magnitudes of the electric field are not same. i.e the net electric field can't be zero. As the magnitude of electric field due to 9Q charge is more than -4Q charge as distance from the charge is less for 9Q.

3) Now for the last part that is on the right of -4Q charge the direction of electric field here again are in opposite directions, and now as we move away from the -4Q charge the magnitude of electric field decreases due to 9Q & -4Q. but this time there will come a point when magnitudes of both the electric field will be same.

This didn't happen on the left of the 9Q as the electric field is given by Kq/r² hence there the charge was also more and distance was also less, effectively giving more electric field due to 9Q.

4) the distance from the charge 9Q let's say 'x'

now equating the magnitude of electric field due to both

K(9Q)/x² = K(4Q)/(x-10)²

9/ x² = 4/ (x-10)²

taking roots on each side:

3/x = 2 / (x-10)

3(x-10) = 2x

3x - 2x = 30

x = 30

x=30m

And lastly the force on any charge is given by product of electric field and its charge, since electric field at point P is zero hence the force on any charge kept at point P will be zero.