Can someone please help me with these physics problems?

Question

a) Two charges A & B are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.179 m from the spot, while charge B is 0.444m from it. Calculate the ratio qB/qA of the charges.

b) Four identical charges (+2.20 micro coulomb each) are brought from infinity & fixed to a straight line. The charges are located 0.416m apart. Determine the electric potential energy of this group.

c) What point charge is at the center of a +24V epuipotential surface that has a radius of 0.199m?

d) The inner and outer surfaces of a cell membrane carry a negative & positive charge, respectively. Because of these charges, a potential difference of about 0.0750V exists across the membrane. The thickness of the membrane is 7.95x10^-9m. What is the magnitude of the electric field in the membrane?

Answer 1

a) equal potentials: k qA / dA = kqB / dB

so qB/qA = dB/dA, the ratio of the distances

b) Use the superposition principle to sum up V = k q^2 / d for each of the six pairs of charges (three that are a distance D apart, two that are a distance 2D apart, and one that is a distance 3D apart)

PE = kq^2 (3/D + 2/(2D) + 1/(3D))

= 13 kq^2 / 3D

c) V = kq/r, so q = Vr/k

d) E = V/d

Answer 2

a)
The potential created by an electric charge (setting infinity as the zero-potential point) is proportional to the amount of charge, and inversely proportional to the distance. Contrast this with the electric field strength being inversely proportional to the distance SQUARED.

By the given information, we know that
Va = Vb
qA/rA = qB/rB
qA/qB = rA/rB
qA/qB = (0.179 m)/(0.444 m)
qA/qB = 0.403

where Va and Vb are potentials, qA and qB are charges, and rA and rB are distances.

Answer 3

D = 55ft. t = sqrt(second/g) t = sqrt(110ft/32ft/s^2) t = a million.86s I easily do no longer comprehend the way you purchased 0.33s. Which in case you paintings it out backwards and also you clean up g. g = second/t^2 g = 1010ft/s^2 Acceleration = speed / Time speed = Distance / Time truly Acceleration = (Distance / Time) / Time = Distance / (Time)^2 aka Distance over seconds squared. So no, you could't overlook concerning the seconds squared even as fixing for time that's squared. so that you may want to sq. root the time with the intention to get the truly fee of time, and by no ability time squared.