Please help In the diagram, q1 is 5.70 nC , q2 is 8.90 nC , q3 is 1.80 nC , and
ID: 1399293 • Letter: P
Question
Please help
In the diagram, q1 is 5.70 nC, q2 is 8.90 nC, q3 is 1.80 nC, and q4 is 5.70 nC. The distance d is 6.30 mm. Find the potential at point A in the diagram, at the center of the square. Assume that V = 0 at infinity.
Find the potential at B.
Find the potential at C.
2.
The electric field strength between two parallel conducting plates separated by 5.00 cm is 7.18×104 V/m. What is the potential difference between the plates?
The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 4.00 cm from the other)?
Explanation / Answer
as there are 2 questions.For first figure is not given .As per guidelines,only 1 can be answered.
so i will solve second one.Please raise different question for other one.
E=7.18×104 V/m.
d=5.00 cm=0.05m
V=Ed
where E is the electric field strength, V is the potential difference and d is the distance between the plate.
V=7.18×104*0.05m
V=0.359*10^4V --answer
V=3.59kV--answer
b)V=Ed
V=7.18×10^4*0.01m
V=0.0718*10^4
V=718V--answer to b