Question

Two charges 5m c & -5mc are placed at points A &
Be which are separated by a distance of 0.06m.
find the electric field at a point P on the
perpendicular bisector of AB at a distance of 0.04
m from its middle point.​

Answer 1

E = -kq/r²

E total at P

= Electric field due to +5mc and Electric field due to -5mc

E due to 5mc =

$\frac{9 \times {10}^{9} \times 5 \times {10}^{ - 9} }{ {0.04}^{2} } = \frac{45}{0.0016} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{450000}{16} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= 28125n {c}^{ - 1}$

E due to -5mc =

$\frac{9 \times {10}^{9} \times 5 \times {10}^{ - 9} }{ {0.04}^{2} } = \frac{45}{0.0016} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{450000}{16} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= 28125n {c}^{ - 1}$

At point P, the +5mc will repel a unit positive charge placed at P and

the -5mc will attract the unit positive charge

Therefore, both the direction of electric fields are in the same direction

Hence E total at P = 28125+28125 = 56250 N/C

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Answer 2

I hope this helps ( ╹▽╹ )