Question

Electric field at point p (2m, 1m,2m) due to a point charge 5.0mu C placed at the origin will be​
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Answer 1

Answer:

The magnitude of the electric field is 9 N/C

Answer 2

Answer:

The electric field is given by 1/3(10$\hat i$ + 5$\hat j$ + 10$\hat k$) × 10³ N/C.

Explanation:

Given,

$\vec{r}$ = 2$\hat{i}$ + $\hat j$ + 2$\hat k$

Charge of point charge object (q) = 5 μC

To find,

The electric field at a point 'r'

Calculation,

The electric field in the vectorial form is given by:

$\vec E = \frac{kq}{|\vec r|^3} \vec r$

Where, k = electrostatic constant = 9 × 10⁹ Nm²C⁻².

Considering the magnitude part of $\vec E$.

Magnitude = kq/r³

But $r = \sqrt{2^2 + 1^2 + 2^2} = 3 m$

⇒ magnitude = (9 × 10⁹ Nm²C⁻²) × (5 × 10⁻⁶ C)/3³

⇒ magnitude = 5/3 × 10³ N/C.

Hence, the electric field $\vec E$ at a point 'r' is   $\frac{5}{3} \times (2\hat i + \hat j + 2\hat k)\times 10^3 N/C$.

Therefore, the electric field is given by 1/3(10$\hat i$ + 5$\hat j$ + 10$\hat k$) × 10³ N/C.

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