Question

Determine the electric field (magnitude and direction) at the point A (8.00 nm, 6.00 nm) caused by a particle located at the origin and carrying a charge of 7.00 μC .

Answer 1

E1 = 9.83 x
`10^(20)`
`NC^(-1)`

E2 = 1.748 x
`10^(21)`
`NC^(-1)`

Step-by-step explanation:

E1 = k Q/r2 = 8.99 x
`10^(9)` x 7 x
`10^(-6)` / 8 x
`10^(-9)` x 8 x
`10^(-9)` = 9.83 x
`10^(20)`
`NC^(-1)`

E2 = k Q/r2 = 8.99 x
`10^(9)` x 7 x
`10^(-6)` / 6 x
`10^(-9)` x 6 x
`10^(-9)` = 1.748 x
`10^(21)`
`NC^(-1)`

The direction of the electric field will be from E1 to E2...