Question

*How much charge will appear on the plates if a capacitor has capacitance of 177 pF when potential difference of 100 V is applied?* 1️⃣ 1.77 nC 2️⃣ .177 nC 3️⃣ 17.7 nC 4️⃣ 17.7 mC​

Answer 1

Given:

Capacitor has capacitance of 177 pF with potential difference of 100 V.

To find:

Charge on the plates?

Calculation:

• We know that the relationship between charge, capacitance and potential difference is as follows:

$Q = C \times V$

$\implies Q = (177 \times {10}^{ - 12} ) \times (100)$

$\implies Q = 177 \times {10}^{( - 12 + 2)}$

$\implies Q = 177 \times {10}^{ - 10}$

$\implies Q = 17.7 \times {10}^{ - 9} \: C$

$\implies Q = 17.7 \: nC$

So, charge on the capacitor plates is 17.7 nC.

Answer 2

Answer:-

• Capacitance =C=177pF
• Potential Difference=V=100V
• Charge=Q=?

we know

$\boxed{\sf Q=CV}$

$\\ \sf\longmapsto Q=177\times 10^{-12}\times 100$

$\\ \sf\longmapsto Q=177\times 10^{-12}\times 10^2$

$\\ \sf\longmapsto Q=177\times 10^{-12+2}$

$\\ \sf\longmapsto Q=177\times 10^{-10}$

$\\ \sf\longmapsto Q=17.7\times 10^{-9}C$

Option C is correct