Question

Consider a square which is 1.0 m on a side. Charges are placed at the corners (marked A,B,C D) of the square as shown in the figure (attached below). Q = 7 μC and q = 2 μC The distance of each corner to the center of the square is 0.707 m.

(Questions below.)

Answer 1

a) What is the magnitude of the electric field at the center of the square?
Express your answer using three significant figures. Answer in N/C.

b) What is the direction of the electric field at the center of the square? (The angle is measured in degrees CCW from the positive x axis in the usual sense.) Answer in degrees.

c) What is the electric potential at the center of the square? Answer in Volts.

d) If a −1μC charge is placed at the center of the square, what is the x-component of the electrostatic force on the charge? Answer in N x-hat.

e) If a −1μC charge is placed at the center of the square, what is the y-component of the electrostatic force on the charge? Answer in N y-hat.

f) If a −1μC charge is placed at the center of the square, how much work (positive or negative) would you have to do to remove it to infinity? Answer in J.

Answer 2

Lets begin from question a

do you know wht is the formula for E field? and then use superposition?

Answer 3

Is that E = kq/r^2?

E = (k)(q1)(q2)(q3)(q4) / r^2
E = (9*10^9)(7*10^-6)(2*10^-6)(7*10^-6)(-2*10^-6) / 0.707^2
E = -3.53*10^-12

I have no idea what the superposition rule is, though.

Answer 4

Oh wait... at the center, electric field due to charges at A & C will cancel out each other due to having equal and opposite field direction. So B & D are the only ones with an impact.

E = kq / r^2 + k |-q| / r^2

= k x 2q / r^2

= 8.98 x 10^9 x 2 x 2 x 10^-6 / (0.71^2)

= 71.26 x 10^3 N/C.

Answer 5

b) Since A & C cancel out, the fields due to μ diagonally opposite 2 μC and -2 μC charges add to each other producing an angle of 135 degree with positive x axis.

^ I remember reading that, so I got b) right too.

Answer 6

c) V = ((2)(9*10⁹)(7*10⁻⁶))/(0.707) = 1.78*10^5 V/m

Okay, got that too.

Answer 7

d) (7.2*10^5)*(1*10^-6)*cos 45
= 0.72 cos 45
= 0.5

Uhhh that was wrong. I know d) & e) will be the same, but I don't know how to do this one.

Answer 8

K*q1*q2/r^2 +k*q1*q2/R^2 ***keep track of the sines. remember one point is pulling the other is pushing.

Answer 9

Oh, so the force acting on the charge will be in the direction along the line joining B & D towards B and will be negative.

F = (71.26 x 10^3) x (1 x 10^-6)
= 71.26 x 10^-3 N Fx
= F x Cos 45
= 50.38 x 10^-3 N x-hat

Looks like I had a sin/cos flip and some exponent issues.

I thought e) would be the same, but apparently it's not?

Answer 10

Oh, wait. It's just -50.38*10^-3.
Got it.

Then... the last one.... f)

Answer 11

if is a lot like the last one, just sum the integrals for each charge.

Answer 12

Er, what's an integral?
This is algebra-based physics and I haven't got to that in calculus yet, so I'm not sure what they are really.

50.38 x 10^-3 + (-50.38*10^-3) = 0
And if I just ignore signs it's 0.1 and both are wrong.

I only have one last chance for f)

:(

Answer 13

yea no integrals required here :P

Answer 14

w=f*d==> q(v)/r q(V(b)-V(a)) be

Answer 15

A group will look like W1=q*V(r) ; W2=k*q2*q1/|r2-r1|, W3=k*q3*(q1/|r3-r1| +q2/|r3-r2|).....see the pattern?

Answer 16

at infinity V always = 0

Answer 17

Total work = w1+w2+w3+w4

Answer 18

Ah, I got it wrong.
Most likely it was an algebra mishap.

Oh well. Thanks for the help, guys!