Two positive charges, each one 10 µC lie on the x axis and form the base of an e
ID: 1992204 • Letter: T
Question
Two positive charges, each one 10 µC lie on the x axis and form the base of an equilateral triangle with sides 20 cm. A negative charge of -2 µC is at the peak of the triangle.(a) What is the x component of the electric field at a point P midway between the two positive charges (assume the positive x-direction is to the RIGHT)?
N/C
(b) What is the y component of the electric field at a point P midway between the two positive charges (assume positive y-direction is UP)?
N/C
(c) If a negative charge of -2 nC is placed at the point P, what is the x-component of the NET force on it?
N
(d) If a negative charge of -2 nC is placed at the point P, what is the y-component of the NET force on it?
N
Explanation / Answer
a)
At A point P midway between the two positive charges we have x component of electric field=0. As electric field of each charge cancels out at the mid point.So x-component =0
b)
In y-direction we have electric field only due to the -ve charge...because the effect of the two positive charges cancel out
distance from the -ve charge =3 a /2
So Electric field = kq/d2 = 9*109 *(10 *10-6)/( 3 20 /200)2
= 3 * 106 N/C ...Direction = UP...so it is positive
c)
x-component of force = 0...as there is no x-component of electric field.
d)
y-component of force = Electric field * charge
= -2 * 10-9 *3 * 106
=-6 * 10-3 N
Direction = Downwards