Question

b) fill-in example: Find the electric field at field point B 2 m 2m 2 m A +C solution: the field directions at B are 'away from +, towards 2 m 2m 2m +++ -21C +C now for the magni tude and the charge is 2uC= 2 x10-6 C, so the distance is klQ E = for each field vector

Answer 1

field of + 2uC

here

r² = 2² + 2² = 8

x - component = 9e9 * 2e-6 * cos 45 / 8 = 1590.99 N/C

y - component = - 9e9 * 2e-6 * sin 45 / 8 = - 1590.99 N/C

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field of - 2 uC

x - component = 9e9 * 2e-6 * cos 45 / 8 = 1590.99 N/C

y - component = - 9e9 * 2e-6 * sin 45 / 8 = 1590.99 N/C

so,

we can see that y - component cancels out

we have only x components which will add up

so,

resultant = y - component = 0

resultant = x - component = 3181.98 N/C

so,

magnitude = 3181.98 N/C

direction = to the right