What is the net force on a small electric dip | vedclass

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(A) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor. The circuit simplifi

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$0$

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(A) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor. The circuit simplifies to a series combination of two $2 \ \Omega$ resistors connected to a $16 \ V$ battery. The current in the circuit is $I = \frac{16 \ V}{2 \ \Omega + 2 \ \Omega} = 4 \ A$. The voltage across the capacitor is equal to the voltage across the $2 \ \Omega$ resistor connected in parallel to it. Thus,$V_c = I 	imes R = 4 \ A 	imes 2 \ \Omega = 8 \ V$. Inside a parallel plate capacitor,the electric field is uniform. The net force on an electric dipole in a uniform electric field is given by $\vec{F} = q\vec{E} + (-q)\vec{E} = 0$. Therefore,the net force on the dipole is $0$.

What is the net force on a small electric dipole placed inside a parallel plate capacitor at steady state,if the plates are separated by $1 \ cm$? The circuit diagram is provided.

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