MHT CET · Physics · Electrostatics
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 \(x=3 \mathrm{~m}\) then, the potential difference \(V_A-V_0\) in volts is
- A -80
- B -220
- C -180
- D -120
Answer & Solution
Correct Answer
(C) -180
Step-by-step Solution
Detailed explanation
Concept: electric field is defined as,
\((-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} l})=\mathrm{d} V\)
Given \(\vec{E}=20 x^2 \hat{i}\) and \(\overrightarrow{\mathrm{d} l}=\mathrm{d} x \hat{i}\)
\(\therefore\left(-20 x^2 \mathrm{~d} x\right)=\mathrm{d} V\)
Integrating between, \(x=0\) to \(3 \mathrm{~m}\)
\(\begin{aligned}
& -\int_0^3 20 x^2 \mathrm{~d} x=\int_{V_0}^{V_A} \mathrm{~d} V \\
& \Rightarrow\left(V_A-V_0\right)=(-20) \times\left.\left(\frac{x^3}{3}\right)\right|_0 ^3 \mathrm{~V}=(-180) \mathrm{V}
\end{aligned}\)
\((-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} l})=\mathrm{d} V\)
Given \(\vec{E}=20 x^2 \hat{i}\) and \(\overrightarrow{\mathrm{d} l}=\mathrm{d} x \hat{i}\)
\(\therefore\left(-20 x^2 \mathrm{~d} x\right)=\mathrm{d} V\)
Integrating between, \(x=0\) to \(3 \mathrm{~m}\)
\(\begin{aligned}
& -\int_0^3 20 x^2 \mathrm{~d} x=\int_{V_0}^{V_A} \mathrm{~d} V \\
& \Rightarrow\left(V_A-V_0\right)=(-20) \times\left.\left(\frac{x^3}{3}\right)\right|_0 ^3 \mathrm{~V}=(-180) \mathrm{V}
\end{aligned}\)
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