A cube is placed inside an electric field, $\vec{E} = 150 y^{2} \hat{j}$ The side of the cube is $0.5 m$ and is placed in the field as shown in the given figure. The charge inside the cube is:

Question

A cube is placed inside an electric field, E → = 150 y 2 j ^ The side of the cube is 0 . 5 m and is placed in the field as shown in the given figure. The charge inside the cube is:

TestHub · Accepted Answer

By gauss law ∮ E → · d s → = q in ε 0 ε 0 ( 150 ) ( 0 . 5 ) 2 × ( 0 . 5 ) 2 = q i n q i n = 8 . 85 × 10 - 12 × 150 16 = 8 . 3 × 10 - 11 C

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