Question

Plates of capacitor carries unequal charges as shown in figure, then which statement is wrong

• A. $V_A>V_B$
• B. $E_A=E_B$
• C. $E_C=0$
• D. $|E_A|>|E_C|$

Answer 1

The correct option is C $E_C=0$

On outside plates of parallel plate capacitors, charge distribute equally.

$8-x=x-6$
8+6 =2x
$\Rightarrow x=7$

Field inside capacitor because of charges on both the plate remains same as there is no charge in between capacitor plate . Therefore, $E_A=E_B$.

Potential across field decreases therefore A has more potential than B.

As both the plates have net charges of opposite sign, electric field inside capacitance are additive in nature while electric field outside capacitance will get substracted . Therefore,
$E_A>E_C$

Also as sum of net charges on both plate is non zero, therefore field will be present both inside and outside the capacitor.Therefore, electric field at point C is non zero.