A1.10 μf capacitor is connected in series with a 1.92 μf capacitor. the 1.10 μf capacitor carries a charge of +10.1 μc on one plate, which is at a potential of 51.5 v. (a) find the potential on the negative plate of the 1.10 μf capacitor. (b) find the equivalent capacitance of the two capacitors.

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

Put the value into the formula

The potential on the second plate

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

Put the value into the formula

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west