Recent content by jelliDollFace

anyone, its been 4 days, please help

how do i factor in the angle into the dipole moment for part a?

Homework Statement what is the potential 18 cm from a dipole moment 2.6nCm at a) 42 deg to axis b) on the perpendicular bisector note: dipole separation << 18cm Homework Equations electric dipole moment, p = qd where q is charge, d is distance electric potential for point...

since the point is now between the charges, the positive points to the right so it is positive, and the negative charge's field is positive direction too

i thought my signs were correct this time? i made them such that they were in the same directions as the first problem, after we corrected them. if i reverse the signs, the magnitude will be same, and wouldn't they just cancel out? and yes i was talking about point A, in the middle when is...

just another quick question. if was to determine the electric field at point P, is this how i would do it. E1 = kq_l/r_1^2 = (9*10^9)(-2*10^-6) / (0.025^2) = -2.88*10^7 n/c E2 = kq_r/r_2^2 = (9*10^9)(+2*10^-6) / (0.025^2) = +2.88*10^7 n/c so Enet = E1 + E2 = 0 so (Ex, Ey = 0, 0) correct...

thanks so much , it worked! it was the sign all along, that's something i need to remember

the positive charge electric field points away from the charge, it should be negative the negative charge electric field points into the charge, it should be positive is i change the signs in my original calculations, will my answer be correct?

i can't see how i got the charges wrong in my calculations...q_l = q_left = left charge = positive charge = +2*10^-6 coulombs, and the distance is 2.5cm from the positive charge, so 0.025m. as for the q_r = q_right = negative charge = -2*10^-6 coulombs, and the distance is 7.5cm from the...

no, the positive charge in the diagram is on the left, not the right. how would having the charge on either side affect the result?

Homework Statement point A is between two charges, one to the left (+2 microcoul) and one to the right (-2 microcoul). point A is midway between the two charges. the two charges are 5cm apart. what is the electric field in the plane of the page (diagram) at a point 5cm left from point A...

thanks so much, it was right!

okay i see now, aside from the issue with the sine and cosine, was my approach for solving for q_red correct? here it is with corrections [(q_red)(+q)k]/d1^2 + [(+q)(-2q)(k)(cos(theta))]/d2^2 = 0 q_red = -[k(+q)(-2q)(d1^2)(cos(theta))]/[(d2^2)(+q)(k)] q_red =...

so this is what i got, since we know the x components must sum to zero soo... [(q_red)(+q)k]/d1^2 + [(+q)(-2q)k]/d2cos(theta)^2 = 0 so q_red = [-k(+q)(-2q)(d1^2)]/[(d2cos(theta)^2)(+q)(k)] is this right?

yayy, thanks so much, the whole 'equal but opposite' thing totally skipped my mind