Question

Two charges are placed as shown in the figure(Figure 1)with q1=3.1?C and q2=?5.0?C.

Two charges are placed as shown in the figure(Figure 1)with q1=3.1?C and q2=?5.0?C.

Find the potential difference between points A and B.

Vab = Va - Vb = ??? V

Answer 1

Va = V1 + V2 = 9.0E9*3.1E-6/0.1 + 9.0E9*-5.0E-6/sqrt(0.1^2 + 0.1^2)=-39198

Vb = 9.0E9*-5.0E-6/0.1 + 9.0E9*3.1E-6/sqrt(0.1^2 + 0.1^2)=-252717

so Va - Vb = -39198 +252717=213519 V

Answer 2

Va = kq1/r1 + kq2/r2

Vb = kq1/r2 + kq2/r1

Va - Vb = k ( q1(1/r1 - 1/r2) - q2(1/r1 - 1/r2))

= k*(q1-q2)(1/0.1 - 1/sqrt(2)*0.1)

= 9*10^3*(8.1)*(10 - 10/sqrt(2))

= 2.135 * 10^5 V

Answer 3

Answer 4

Since the q3 is negative, it will be repelled by q2 and it will be attracted by q1. In zone A, the repulsion of the force between q2 and q3 is stronger than the attraction of the force between q1 and q3, therefore, it cant be in zone A

in zone B, it can't work, since the repulsion and the attraction are in the same direction... which would modify the whole system.

this leaves us with zone C, where you can have the repulsion (F1) equal the attraction (F2).

Now that we have conceptualize, we can gather the formula for this problem, and we only need one:

kqq

We want F1 to equal F2, so:

(note that for this kind of problem, you need a relation between the "r" in the formula, which is why you use one x and the other is simply x + the distance between the two)

x +0.2)

Because we have k and q3 on both side, we can clear them out, leaving us with:

x2(0.2)2

simple algebra gives you:

qi (x+0.2)-= q2 (x).

instead of opening that equation, let's instead use a square root on both sides:

or

i 0.2 gix q | 0.2 =

Now, let's substitute:

V3.2 × 10-6 x +V32 × 10-6 0.2-v-8.3 × 10-6 X

ANSWER:

0.001789x + 0.0003577 = 0.002881x

0.0003577 = 0.001092x

(0.0003577)/(0.001092) = x

x = 0.33m to the right of q1