Question

At point A (0,0,4) a positive charge Q is located and at point B (0,0,-4), a negative charge of magnitude Q is located. Calculate the direction of the electric field intensity at point C situated at (5,0,0,).A) -Z directionB) +Z direction C) - X directionD) -Y direction

Answer 1

Given info : At point A (0,0,4) a positive charge Q is located and at point B (0,0,-4), a negative charge of magnitude Q is located.

To find : Calculate the direction of the electric field intensity at point C situated at (5,0,0,).

Solution : see figure,

magnitude of electric fields for both charge at O are same. I.e., |E|= KQ/|r|²

Where |r| = √(5² + 4²)

We have to get direction of each electric field.

for charged particle placed at B.

E₁ = |E|$\hat{BO}$

= |E| (5i - 4k)/|r|

Similarly, for charged particle placed at A,

E₂ = |E|$\hat{OA}$

= |E|(-4k - 5i)/|r|

Net electric field at O, E = E₁ + E₂

= |E|(5i - 4k)/|r| + |E|(-4k - 5i)/|r|

= |E|/|r| (5i - 4k - 4k - 5i )

= |E|/|r| (-8k)

Hence it is clear that net electric field acts in negative z - axis (-k). So the correct option is (A)

Answer 2

Answer:

a

Explanation:

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