A cube with sides of area 32 cm2 contains a 35.9 nanoCoulomb charge. Find the flux of the electric field through the surface of
the cube in unis of Nm2/C.
1 answer:
Answer:
The electric flux through the surface is equal to 3.878 x 10³ Nm²/C
The field distance r is equal to half the length of each side of the cube.
From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C
The total flux area available to this electric field is 6x32cm²
Explanation:
The full solution can be found in the attachment below.
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Answer:
Explanation:
Given
Force of repulsion between two charge particle is given by force F
Electrostatic force is given by
where and
is the charges of particle
r=distance between charge particle
when charges are doubled and distance is reduced to half
i.e. q become 2 q and r becomes 0.5 r