Consider two point charged objects separated by a distance of 1.72 cm. The charg
ID: 251662 • Letter: C
Question
Consider two point charged objects separated by a distance of 1.72 cm. The charge on the left hand object is +30.5 C. The charge on the right hand object is -30.5 C. What is the strength (magnitude) of the electric field due to the left hand object only at the location exactly halfway in between the two charged objects? Tries 0/10 What is the strength (magnitude) of the electric field due to the right hand object only at the location exactly halfway between the two charged objects? Tries 0/10 What is the strength of the net electric field due to the BOTH objects at the point halfway between the two charged objects? Tries 0/10 What would be the force on a proton placed at the point halfway between the objects?
Explanation / Answer
We know that Elecrtic field due to a point charge is kq/r^2 where k = (1/ 4 * pi * epsilon) which is equal to 9 *10^9
q is the charge on the object and r is the distance at which it exerts its fiesld strength.
So Vectorially Let us assume both charges are seperated on x axis by 1.72 cm.
Exact midway = 1.72/2 = 0.86 cm
So E due left hand side is 9 * 10^9 * 30.5 * 10^-6 /( 0.86*10^-2)^2 = 3.712 * 10^9 N/C
Since it is a positive charge it is along positive x direction.
(b) In the second case from the charge on right side, q = -30.5 micro columb
distance is same as 1.72/2 = 0.86 cm
So the magnitude of field is also same kq/r^2 = 3.712*10^9 N/C
Since it is a negative charge, direction will be into the charge. Since it is on right side, Let us say on right side of origin, The direction into it means positive x direction.
(c) Since Both charges exert same field in same direction. Net field at the middle point = 2 *3.712 * 10^9 = 7.424 *10^9 N/C
Force on any charge placed on that point experience F = q *E net
So charge of proton = 1.6 * 10^-19 C
So F = 1.6 * 10^-19 * 7.424 *10^9 N = 1.18 *10^-9N