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Physics Question on electrostatic potential and capacitance

In a region, the potential is represented by V(x, y, z) = 6x - 8xy - 8y + 6yz, where VV is in volts and x,y,zx, y, z are in metres. The electric force experienced by a charge of 22 coulomb situated at point (1,1,1)(1, 1, 1) is

A

65 N6 \sqrt 5 \,N

B

30 N

C

24 N

D

435 N4 \sqrt 35 \,N

Answer

435 N4 \sqrt 35 \,N

Explanation

Solution

Here, V(x, y, z) = 6x - 8xy - 8y + 6yz
The x, y and z components of electric field are
Ex=−∂V∂x=−∂∂x(6x−8xy−8y+6yz)E_x = - \frac{\partial V}{\partial x} = - \frac{\partial}{\partial x} (6x - 8xy - 8y + 6yz)
      =−(6−8y)=−6+8y\, \, \, \, \, \, = -(6 - 8y) = -6 + 8y
Ey=−∂V∂y=−∂∂y(6x−8xy−8y+6yz)E_y = - \frac{\partial V}{\partial y} = - \frac{\partial}{\partial y} (6x - 8xy - 8y + 6yz)
       =−(−8x−8+6z)=8x+8−6z\, \, \, \, \, \, \, = -(-8x - 8 + 6z) = 8x + 8 - 6z
Ez=−∂V∂z=−∂∂z(6x−8xy−8y+6yz)=−6yE_z = - \frac{\partial V}{\partial z} = - \frac{\partial}{\partial z} (6x - 8xy - 8y + 6yz) = - 6y
E→=Exi^+Eyj^+Ezk^\overrightarrow {E} = E_x \widehat {i} + E_y \widehat {j} + E_z \widehat {k}
=(−6+8y)i^+(8x+8−6z)j^−6yk^= (-6 + 8y) \widehat {i} + (8x + 8 - 6z) \widehat{j} - 6y \widehat {k}
At point (1, 1, 1)
E→=(−6+8)i^+(8+8−6)j^−6k^\overrightarrow {E} = (-6 + 8) \widehat {i} + (8 + 8 - 6) \widehat {j} - 6 \widehat {k}
=2i^+10j^−6k^= 2 \widehat {i} + 10 \widehat {j} - 6 \widehat {k}
The magnitude of electric field E→\overrightarrow {E} is
E→=Ex2+Ey2+Ez2=(2)2+(10)2+(−6)2\overrightarrow {E} = \sqrt {E_x^2 + E_y^2 + E_z^2} = \sqrt {(2)^2 + (10)^2 + (-6)^2}
=140=235 N C−1= \sqrt 140 = 2 \sqrt 35\, N\, C^{-1}
Electric force experienced by the charge
F=qE=2 C×235 N C−1=435 NF = qE = 2 \,C \times 2 \sqrt 35\, N \,C^{-1} = 4 \sqrt 35\, N