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.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards