The two charges in the figure below are separated by d 3.00 cm. (Let q1 18.5 nc
ID: 1526049 • Letter: T
Question
The two charges in the figure below are separated by d 3.00 cm. (Let q1 18.5 nc and q2- 29.5 nc.) 60.0 degree (a) Find the electric potential at point A. (b) Find the electric potential at point B, which is halfway between the charges. Consider the Earth and a cloud layer 500 m above the planet to be the plates of a parallel-plate capacitor. (a) If the cloud layer has an area of 4.0 km^2 4000000 m^2, what is the capacitance? (b) If an electric field strength greater than 3.0 times 10^6 causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold.Explanation / Answer
a) To find the electric potential at point A, we must find the potential due to each charge individually, then just add them. So, using Coulomb's law for potential we have
P(q1) = K q1 / r J
where K = Coulomb's constant = 9 * 10^9 ; r = 0.03 m
P(q2) = K q2 / r J
P(A) = P(q1) + P(q2) = (K/r) ( q1 + q2)
P(A) = 9 * 10^9 [(-18.5 * 10^-9) + (29.5 ^ 10^-9)] / 0.03
P(A) = 9 * 10^9 ( 11 * 10^-9) / 0.03
P(A) = 3.30 kJ
b) We have an isoceles triangle, therefore the base angles are equal. The angle at A is (180 - 120) = 60 degrees. So we now know that it is an equilateral triangle. So the distance between the charges must be 3.0 cm or 0.03 m.
Therefore, the half way point is 0.03/2 = 0.015 m
P(B) = (K/r) (q1 + q2)
where, this time, r = 0.015 m
P(B) = (9 * 10^9 / 0.015) ( 11 * 10^-9) = 6.6 kJ
2.
(a).
C = K Ao/d
C = (1)(4000000)(8.85 * 10^(-12))/500
C = 7.08 * 10^(-8) Farad
(b).
V = E d
Q = C V
Q = C E d
Q = (7.08 * 10^(-8))(3.0 * 10^6)(500)
Q = 106.2 Coulomb
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.