The two charges in the figure below are separated by a distance d = 5.00 cm, and
ID: 2171043 • Letter: T
Question
The two charges in the figure below are separated by a distance d = 5.00 cm, and Q = +7.30 nC.(a) Find the electric potential at A.
1 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kV
(b) Find the electric potential at B.
2 .
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kV
(c) Find the electric potential difference between B and A.
Explanation / Answer
In this problem you basically just sum up the potentials acting on each point. Using the formula V = sum(k*qi/ri) k = 9x10^9 (a) First for Va Va = (kQ/d) +(k(2Q)/sqrt(d^2+d^2)) notice that 2Q is 2*Q so 2*7.50x10^-9 for the distance for the second sum, that was just pythagorean theorem a^2+b^2 = c^2 Inputing the numbers for a we have: (9x10^9)*(7.50x10^-9)/(1.50x10^-2) + (9x10^9)*(2*7.50x10^-9)/sqrt((1.50x10-2)… Va = 4500 + 6363.96 = 10863.96V or 10.86kV (b) For b its the exact same process, except the distances are different for the charges. Vb = (kQ/sqrt(d^2+d^2)) +(k(2Q)/d) (9x10^9)*(7.50x10^-9)/sqrt((1.50x10-2)… + (9x10^9)*(2*7.50x10^-9)/(1.50x10-2)) 3181.98 + 9000 = 12181.98V or 12.18kV (c) The difference is just simply Vb - Va: 12181.98 - 10863.96 = 1318.02V Hope that helps, sorry if I made mistakes, this is the easiest way to input formulas.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.