how do i solve the following
2 opposite charges of the same magnitude 2*10^-7 C are separated by a distance of 15cm.
a) what is the magnitude and direction of the electric field strength at the midpoint between the charges?
b)what is the magnitude and direction of the force acting on an electron placed at the midpoint?
Thanks
(a) The resultant electric field at a point due to a system of point charges is given by the vector sum of the electric field intensities due to the individual charges.
Let Q
1 = 2x10
-7 C and Q
2 = -2x10
-7 C
Electric field strength due to Q
1 = (2x10
-7)/(4 x 22/7 x (8.85x10
-12) x (15x10
-2)/2) = 2.40 x 10
4 Vm
-1Electric field strength due to Q
2 will have same magnitude and in same direction since they are oppositely charged.
Resultant electric field intensity will be 2(2.40x10
4) =
4.80x104 Vm
-1The direction of the electric field is always along the direction of decreasing electric potential. Therefore from the positive charge to the negative one.
(b) Use the formula F = qV/d
Force = (2.0x10
-7)x(4.80x10
4)/((15x10
-2)/2) =
6.39x10-2 N
The direction of the force towards the positive charge, that is against the direction of the electric field.
Hope it helps