A cube is placed inside an electric field, The side of the cube is 0.5m and is placed in the field as shown in the given figure. The change inside the cube is:

Option 1 -

8.3 * 10^-11C

Option 2 -

8.3 * 10^-12C

Option 3 -

3.8 * 10^-12C

Option 4 -

3.8 * 10^-11C

12 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 1

Detailed Solution:

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m So

$E = 150 \times {(0.5)}^{2} = \frac{150}{4}$

$f l u x f l o w i n g ? = E A = \frac{150}{4} \times {(0.5)}^{2}$

$= \frac{150}{16}$

Gausses law $? = \frac{q}{ε_{0}}$

$\frac{150}{16} = \frac{q}{ε_{0}}$

$= 8.3 \times 1 0^{- 11} C$

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A cube is placed inside an electric field, The side of the cube is 0.5m and is placed in the field as shown in the given figure. The change inside the cube is:

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