Question

the two charges are separated by a distance of 2a and point p is at a distance x from the midpoint between the two charges (figure 1), q>0
figure
1 of 1
part b
determine the direction of the electric field at the point p
to the right
to the left
upward
downward

Explanation:

Step1: Analyze electric field due to +Q

The electric field due to a positive charge \(+Q\) at point \(P\) is directed away from \(+Q\). Using Coulomb's law, the electric field \(E_{+Q}\) due to \(+Q\) at \(P\) has a magnitude \(E_{+Q}=\frac{kQ}{(a + x)^2}\) and is directed to the right.

Step2: Analyze electric field due to -Q

The electric field due to a negative charge \(-Q\) at point \(P\) is directed towards \(-Q\). The magnitude of the electric field \(E_{-Q}\) due to \(-Q\) at \(P\) is \(E_{-Q}=\frac{kQ}{(x - a)^2}\) (since \(x> a\) as per the problem setup where \(P\) is at a distance \(x\) from the mid - point and the charges are separated by \(2a\)), and is directed to the right (because the field due to a negative charge is towards the charge, and since \(P\) is to the right of \(-Q\), the field due to \(-Q\) at \(P\) is to the right).

Answer:

to the right