Two conductors having net charges of +18.0 µC and -18.0 µC have a potential difference of 18.0 V between them.(a) Determine the capacitance of the system. Check the units of capacitance and potential difference carefully. F(b) What is the potential difference between the two conductors if the charges on each are increased to +324.0 µC and -324.0 µC?
Question
Two conductors having net charges of +18.0 µC and -18.0 µC have a potential difference of 18.0 V between them.(a) Determine the capacitance of the system. Check the units of capacitance and potential difference carefully. F(b) What is the potential difference between the two conductors if the charges on each are increased to +324.0 µC and -324.0 µC?
Solution
(a) The capacitance (C) of a system can be determined using the formula:
C = Q/V
where Q is the charge and V is the potential difference. In this case, Q = 18.0 µC = 18.0 x 10^-6 C (converting microcoulombs to coulombs) and V = 18.0 V.
So,
C = (18.0 x 10^-6 C) / 18.0 V = 1.0 x 10^-6 F = 1.0 µF
Therefore, the capacitance of the system is 1.0 µF.
(b) If the charges on each conductor are increased to +324.0 µC and -324.0 µC, the potential difference (V) can be determined using the formula:
V = Q/C
where Q = 324.0 µC = 324.0 x 10^-6 C and C = 1.0 µF = 1.0 x 10^-6 F.
So,
V = (324.0 x 10^-6 C) / (1.0 x 10^-6 F) = 324.0 V
Therefore, the potential difference between the two conductors when the charges on each are increased to +324.0 µC and -324.0 µC is 324.0 V.
Similar Questions
Two capacitors (C1 = 8.00 µF and C2 = 13.0 µF) are now connected in series and to a 9.00-V battery.(a) Find the equivalent capacitance of the combination. µF(b) Find the potential difference across each capacitor.V1 = What is the relationship between the charges on each capacitor and the charge that would exist on an equivalent capacitor? VV2 = V(c) Find the charge on each capacitor.Q1 = µCQ2 = µC
Two charges are set on the x-axis 11.4 cm away from each other. The charges are -6.10 nC and 21.4 nC. Calculate the electric potential at the point on the x-axis where the electric field due to these two charges is zero.
Three capacitors of capacitance C1 = 3.85 𝜇F, C2 = 2.20 𝜇F, and C3 = 4.95 𝜇F are connected in series. A potential difference of ΔVb = 84.0 V is maintained by a battery.Find the equivalent capacitance of the series of capacitors and the charge on each capacitor.Ceq = 𝜇FQ = 𝜇CDetermine the effect on the equivalent capacitance of reducing the second capacitance to 0.1 times its previous value.Ceq
Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 pF.(ii) Calculate the potential difference between the plates of X and Y.(iii) Estimate the ratio of electrostatic energy stored in X and Y
A positive charge Q1 = +4.80 nC is located at x1 = -2.00 m, a negative charge Q2 = -6.30 nC is located at x2 = 3.00 m, and a positive charge Q3 = 5.40 nC is located at x3 = 9.00 m. Find the net electric potential at x = 0.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.