Question

Electric field at x=10cm is 100v/m and at x=-10 cm is - 100v/m. the magnitude of the charge enclosed by the cube of side 20cm is

Answer 1

$e = \frac{200}{20} = 5 \frac{newton}{coulomb} \\ \\ since \: \: e = \frac{1}{4\pi \: epsilon} \times \frac{q}{ {r}^{2} } \\ \\ \: \: q = \frac{e \times 400}{9 \times {10}^{9} } = \frac{5 \times 400}{9 \times {10}^{9} } coulomb \\ i \: hope \: you \: will \: understood$

Answer 2

Answer:

ans 8Eo I hope this helps you guys