For the potential $V = 3 x ^ { 2 } z - 2 y z ^ { 3 }$ ,what is the corresponding electric field at the point (2,2,2)?
A) $- 24 \tilde { \mathbf { i } } + 16 \tilde { \mathbf { j } } + 36 \tilde { \mathbf { k } }$
B) $12 \tilde { \mathbf { i } } - 16 \tilde { \mathbf { j } } + - 16 \tilde{ \mathbf { k } }$
C) $24 \tilde { \mathbf { i } } - 16 \tilde{ \mathbf { j } } - 36 \tilde { \mathbf { k } }$
D) $- 6 \tilde { \mathbf { i } } + 2 \tilde { \mathbf { j } } + 6 \tilde { \mathbf { k } }$
E)The correct answer is not given.

Question

Quizplus · Accepted Answer

The electric field is given by the negative gradient of the potential:$$\mathbf{E} = -
abla V = - \frac{\partial V}{\partial x} 	ilde{\mathbf{i}} - \frac{\partial V}{\partial y} 	ilde{\mathbf{j}} - \frac{\partial V}{\partial z} 	ilde{\mathbf{k}}$$So, for the given potential, we have:$$\mathbf{E} = - \frac{\partial}{\partial x} (3 x^2 z - 2 y z^3) 	ilde{\mathbf{i}} - \frac{\partial}{\partial y} (3 x^2 z - 2 y z^3) 	ilde{\mathbf{j}} - \frac{\partial}{\partial z} (3 x^2 z - 2 y z^3) 	ilde{\mathbf{k}}$$$$= - (6xz) 	ilde{\mathbf{i}} - (-2z^3) 	ilde{\mathbf{j}} - (3x^2 - 6yz^2) 	ilde{\mathbf{k}}$$$$= -24 	ilde{\mathbf{i}} + 16 	ilde{\mathbf{j}} + 36 	ilde{\mathbf{k}}$$Therefore, the electric field at the point (2,2,2) is:$$\mathbf{E} = -24 	ilde{\mathbf{i}} + 16 	ilde{\mathbf{j}} + 36 	ilde{\mathbf{k}}$$

For the Potential V=3x2z−2yz3V = 3 x ^ { 2 } z - 2 y z ^ { 3 }V=3x2z−2yz3

For the Potential $V = 3 x ^ { 2 } z - 2 y z ^ { 3 }$