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Clf the potential in the region of space around the point (-1 m, 2 m, 3 m) given by V=10 x^2+5 y^2-3 z^2 volts. Calculate the three components of the electric field at this point.

Option: 1

18 \mathrm{v} / \mathrm{m}, 18 \mathrm{v} / \mathrm{m}, 18 \mathrm{v} / \mathrm{m}


Option: 2

10 \mathrm{v} / \mathrm{m}, 20 \mathrm{v} / \mathrm{m}, 25 \mathrm{v} / \mathrm{m}


Option: 3

20 \mathrm{v} / \mathrm{m}, 20 \mathrm{v} / \mathrm{m} 18 \mathrm{v} / \mathrm{m}


Option: 4

20 \mathrm{v} / \mathrm{m}, 12 \mathrm{v} / \mathrm{m}, 18 \mathrm{v} / \mathrm{m}


Answers (1)

best_answer

To calculate the three components of the electric field at the given point (-1 m, 2 m, 3 m), we need to take the negative gradient of the potential function V(x, y, z). The negative gradient of V gives us the electric field vector E

The electric field is given by: 

E=-\nabla V

where \nabla is the del operator, which is a vector differential operator that represents the gradient.
The gradient of a scalar function V=10 x^2+5 y^2-3 z^2 is given by:


\nabla V=(\partial V / \partial x) \hat{i}+(\partial V / \partial y) \hat{\jmath}+(\partial V / \partial z) K

Taking the partial derivatives of  \mathrm{V} with respect to x, y , and z , we get: 

\begin{aligned} & \partial V / \partial x=20 x \\ & \partial V / \partial y=10 y \\ & \partial V / \partial z=-6 z \end{aligned}

Substituting these values into the expression for the electric field, we have: 

E=-[(20 x) \hat{i}+(10 y) \hat{\jmath}+(-6 z) k]

Now, we can substitute the coordinates of the given point (-1 m, 2 m, 3 m) into the expression for the electric field:

E=-(-20 \times 1 \hat{i}+20 \times 1 \hat{j}-6 \times 3 \hat{k})

Simplifying, we get: 

E=-(-20 \hat{i}+20 \hat{j}-18 \hat{k})

Therefore, the three components of the electric field at the point (-1 m, 2 m, 3 m) are: 

\begin{aligned} &E x=-20 \mathrm{v} / \mathrm{m}\\ &\text { Ey }=20 \mathrm{v} / \mathrm{m}\\ &E z=18 \mathrm{v} / \mathrm{m} \end{aligned}

Posted by

Rakesh

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