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 &#8594; = 150 y 2 j ^ The side of the cube is 0 . 5 &#160; m and is placed in the field as shown in the given figure. The charge inside the cube is:

Aicharya · Accepted Answer

By gauss law &#8750; E &#8594; &#183; d s &#8594; = q in&#160; &#949; 0 &#949; 0 ( 150 ) ( 0 . 5 ) 2 &#215; ( 0 . 5 ) 2 = q i n q i n = 8 . 85 &#215; 10 - 12 &#215; 150 16 = 8 . 3 &#215; 10 - 11 C

A cube is placed inside an electric field, E → = 150 y 2 j ^ The side of the cube is 0 . 5   m…

Solution

A cube is placed inside an electric field, E &#8594; = 150 y 2 j ^ The side of the cube is 0 . 5 &#160; m…

Solution

A cube is placed inside an electric field, E → = 150 y 2 j ^ The side of the cube is 0 . 5 m…