ABC is an equilateral triangle of side 0.1m. Point charges of +3nC and -3nC are placed at the corners A & B respectively. Calculate the resultant electric field at C

Answer 1

Q at A and Q at B

add components along axis of altitude of triangle from A B
E along axis = (k 3nC cos 30 - k 3nC cos 30)/.1^2
which is zero
E perpendicular to that axis, in other words parallel to AB is
E = (k 3nC sin30 + k 3nC sin 30)/.1^2
= k *2*.5 *3nC /.01
= 100 k *3nC

Answer 2

The correct answer is given:r=0.1m,q=3nC, we have E=-1×P÷4×pai×apsilon°×rpower3 E=-9×10^9×2×3×10^-9×0.05÷0.1 E=-27×10^2 N÷C

Answer 3

ok ok