Question

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. 'V'. If one of the capacitor is inserted in liquid of dielectric constant 'K', then potential difference of the other capacitor will become ______.

Options

• `("K"+1)/"KV"`

• `"K"/("V"(1-"K"))`

• `"K"/("V"("K"+1))`

• `"KV"/("K"+1)`

Answer 1

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. 'V'. If one of the capacitor is inserted in liquid of dielectric constant 'K', then potential difference of the other capacitor will become `underline("KV"/("K"+1))`.

Explanation:

Let the two identical air capacitors be:

• Initially, both have capacitance C
• Total voltage = V
• One of them is filled with dielectric of constant K, so its new capacitance becomes KC

When connected in series:

• The charge on both capacitors is the same.
• The total potential difference is shared according to the inverse ratio of their capacitances.

Let the potential difference across:

• the capacitor with dielectric = V_i
• the other capacitor (without dielectric) = V₂

Then: `V_1 = Q/(KC), V_2 = Q/C`

Total voltage: `V = V_1 + V_2 = Q/(KC) + Q/C + (Q(1 + K))/(KC) => Q = (VKC)/(K + 1)`

Now, `V_2 = Q/C = (VKC)/(K + 1)·1/C = (KV)/(K + 1)`

Hence, the potential difference across the other capacitor is `(KV)/(K + 1)`.