How Does a Dielectric Constant Affect Capacitor Charge?

[logearav]

Homework Statement

A Parallel Plate Capacitor connected to a 12 V source is charged to 21 micro coulomb. If the capacitor is filled with an oil of dielectric constant 3, then the charge stored is ---

Homework Equations

The Attempt at a Solution

C = q / V, substituting the values i got 1.75 micro farad. Now from the equation C_dielec/ C_air= epsilon_r i now got 5.25 micro farad.
Again C_dielec = q /V, i cross multiplied 12 volt with 5.25 micro farad and got the answer 63 micro coulomb as the answer. But the answer as given in the book is 57 micro coulomb. Is there any mistake in my steps? Please help

 

[anathema]

Hello,

Your approach seems correct, but there may be a small calculation error. Here is how I would solve the problem:

1. Use the formula C = q / V to find the initial capacitance: C = (21 microC) / (12 V) = 1.75 microF.

2. Use the formula C = C0 * k (where C0 is the initial capacitance and k is the dielectric constant) to find the new capacitance after the oil is added: C = (1.75 microF) * 3 = 5.25 microF.

3. Use the formula q = CV to find the final charge: q = (5.25 microF) * (12 V) = 63 microC.

I hope this helps. If you still get a different answer, please double check your calculations. Good luck!