(B) The electric field $E$ between two parallel plates is given by the formula $E = \frac{|\Delta V|}{d}$,where $\Delta V$ is the potential difference

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(B) The electric field $E$ between two parallel plates is given by the formula $E = \frac{|\Delta V|}{d}$,where $\Delta V$ is the potential difference and $d$ is the distance between the plates.Given:Potential difference $\Delta V = V_2 - V_1 = 30 \,V - (-10 \,V) = 40 \,V$.Distance $d = 2 \,cm = 2 \times 10^{-2} \,m$.Substituting the values:$E = \frac{40 \,V}{2 \times 10^{-2} \,m} = 20 \times 10^2 \,V/m = 2000 \,V/m$.Therefore,the correct option is $B$.

Class 12 Physics Electrostatic Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

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Two metal plates are separated by $2 \,cm$. The potentials of the plates are $-10 \,V$ and $+30 \,V$. The electric field between the two plates is: (in $\,V/m$)

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A
$1000$
B
$2000$
C
$3000$
D
$4000$

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Electrostatic Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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Two metal plates are separated by $2 \,cm$. The potentials of the plates are $-10 \,V$ and $+30 \,V$. The electric field between the two plates is: (in $\,V/m$)

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The potential $V$ varies with $x$ and $y$ as $V = \frac{1}{2}(y^2 - 4x) \text{ volts}$. The electric field at $(1 \text{ m}, 1 \text{ m})$ is:

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Assume that an electric field $E=20 x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x=3 \ m$,then the potential difference $V_A-V_0$ in volts is

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