Question:
The electric potential V at any point $(x, y, z)$ in space is given by $V = 3x^{2}$ where $x, y, z$ are all in metre. The electric field at the point (1 m, 0, 2 m) is
The electric potential V at any point $(x, y, z)$ in space is given by $V = 3x^{2}$ where $x, y, z$ are all in metre. The electric field at the point (1 m, 0, 2 m) is
Updated On: Jun 4, 2024
- $6 \,V\, m^{-1}$ along negative $x$-axis
- $6 \,V \,m^{-1}$ along positive $x$-axis
- $12\, V\, m^{-1}$ along negative $x$-axis
- $12\, V \,m^{-1}$ along positive $x$-axis
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The Correct Option is A
Solution and Explanation
Electric potential $V=3 x^{2}$
$E=-\frac{d V}{d x}$
$E=\frac{d}{d x}\left(3 x^{3}\right) $
$E=-6 x $
At the point $(1,0,2)$
Electric field $E=6 \times 1=-6 V / m$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).