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Question

Physics Question on electrostatic potential and capacitance

The potential of the electric field produced by a point charge at any point (x,y,z)(x, y, z) is given by V=3x2+5V= 3x^{2} + 5 , where x,y,zx, y, z are in metres and VV is in volts. The intensity of the electric field at (2,1,0)(-2,1,0) is

A

+17Vm1+17 \,V\,m^{-1}

B

17Vm1-17\,V\,m^{-1}

C

+12vm1+12\,v\,m^{-1}

D

12Vm1-12\,V\,m^{-1}

Answer

+12vm1+12\,v\,m^{-1}

Explanation

Solution

Potential (V)=3x2+5( V )=3 x ^{2}+5
Intensity of the electric field, E=dVdxE=\frac{-d V}{d x}
E=d(3x2+5)dx=6x\therefore E =-\frac{ d \left(3 x ^{2}+5\right)}{ dx }=-6 x
Ex=2=6(2)=12V/m\left.\therefore E \right|_{ x =-2}=-6(-2)=12 V / m