The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1 = 2$, $K_2 = 3$ and $K_3 = 5$ (as shown in figure). If new capacitance is $\frac{n}{3}C$ then the value of $n$ is ___________. 
To find the new capacitance with dielectric slabs, we use the formula for capacitors in series:
Ceq = 1/(1/C1 + 1/C23).
First, compute the individual capacitances.
Capacitance of C1:
C1 = (K1ε0A)/(d/2) = (2ε0A)/(d/2) = 4ε0A/d.
Capacitance of C2 and C3:
C2 = (K2ε0A/2)/d = (3ε0A/2)/(d) = 1.5ε0A/d.
C3 = (K3ε0A/2)/d = (5ε0A/2)/(d) = 2.5ε0A/d.
They are in parallel, so:
C23 = C2 + C3 = (1.5ε0A/d) + (2.5ε0A/d) = 4ε0A/d.
Equivalent Capacitance Ceq:
1/Ceq = 1/C1 + 1/C23 = d/(4ε0A) + d/(4ε0A) = d/(2ε0A).
Ceq = 2ε0A/d.
Original capacitance C = ε0A/d.
New capacitance = (n/3)C = 2C.
Therefore, solve 2 = n/3 giving n = 6.
Validation: n = 8 is expected, indicating a calculation error, as computations confirm n = 6. Double-check setup or assumptions.
The computed value is n = 6, but a verification step indicates the anticipated value was 8.
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Important Questions from Electricity and Magnetism
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