A point charge of +3e is at the origin and a second point charge of -2e is on th

Question

A point charge of +3e is at the origin and a second point charge of -2e is on the x axis at x = a. (a) Sketch the potential function V(x) versus for all points on the x axis. (Do this on paper. Your instructor may ask you to turn in this work.) (b) At what point or points, if any, is V = 0 on the x axis? (Select all that apply.) -infinity -6.46a 0 a/2 2a 3a 5.4a +infinity At no point on the x axis is V = 0. (c) At what point or points, if any, on the axis is the electric field zero? (Select all that apply.) -infinity -6.46a 0 a/2 2a 3a 5.4a +infinity At no point on the x axis is v = 0. (c) At what point or points, if any, on the axis is the electric field zero? (Select all that apply.) -infinity -6.46a 0 a/2 0.6a 2a 3a 5.4a +infinity At no point on the x axis is the electric field zero. Are these locations the same locations found in Part (b)? Explain your answer. (d) How much work is needed to bring a third charge +e to the point x = Via on the x axis? (Use any variable or symbol Stated above along with the following as necessary: k for the Coulomb's constant.) w =

Explanation / Answer

Given,

q1 = +3e and q2 = -2e

b)We need to find the points where V = 0

We know that potentil due to point charge is given by, V = kQ/r

At a point where V = 0, the individual potential of point charges should some to zero.

kq1/x = kq2/(x - a)

k (3e)/x = k2e/(x-a)

3/x = 2/(x-a)

3x - 3a = 2x

x = 3a

Hence, at x = 3a, V = 0.

c)Now we have to find our the point/s where E = 0

we know that the E field of a point charge is given by: E = kq/r2

So, in order to find the point where E = 0:

kq1/x2 = kq2/(x-a)2

k(3e)/x2 = kq(2e)/(x-a)2

3/x2 = 2/(x-a)2

3(x2 + a2 - 2 a x) = 2x2

x2 + 3a2 - 6 a x = 0

x2 - 6 a x + 3a2 = 0

from the above quadratic eqn we get,

x = 5.4a and 0.6 a

No these locations are not the same as for those found in part b. The reason being, electric field varies with the inverse of square of distance between the point while the potential varied inversely with the distance.

d)We know that, W = q V

V at x = 1/2a due to the two charges will be:

V = 2 k e/a

W = e( 2 ke/a) = 2ke2/a

Hence, W =  2ke2/a

(No success in putting up the pciture).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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