Two positive charges, each one 10 µC lie on the x axis and form the base of an e

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.