Question

The electrostatic potential existing in space is given by V = (xy + yz + zx) volts, where x, y and z are in metres. The angle made by electric field at (2 m, 2 m, 2 m) with positive x axis is a. Then

Answer 1

Answer:

cos alpha =√-1/3

Explanation:

hope it will help u

Answer 2

The angle is cos⁻¹ ( -1/1$\sqrt{3}$).

Given

• The electrostatic potential existing in space (V)  = (xy + yz + zx) volts,  

                                           (x, y and z are in metres)

To find

The angle made by the electric field at (2 m, 2 m, 2 m) with positive x  axis

Step by step calculations:

Electric field = E = -ΔV

                            = -( δV/ δx î + δV/ δx j^ + δV/ δx k^)

                            = -(y +z) î  - (x+z) j^ - (x-y) k^

Given

i^ = 2m

j^ = 2m

k^ = 2m

Putting these values in the above equation, we get

= -(y +z) î  - (x+z) j^ - (x-y) k^

= -(2 + 2) î  - (2 + 2) j^ - (2 + 2) k^

= -4i^ - 4j^ - 4k^

E = $\sqrt{(-4^{2}) +(-4^2) +(-4^{2)$

IEI = $\sqrt{48}$

IEI = $4\sqrt{3}$

The angle $\alpha$ made with positive x axis =

 cos $\alpha$ = -4 / 4$\sqrt{3}$

→ cos $\alpha$ = -1/$\sqrt{3}$

→ $\alpha$ = cos⁻¹ ( -1/1$\sqrt{3}$)

Thus, the answer required = cos⁻¹ ( -1/1$\sqrt{3}$)

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