Two point charges, q1 -25 µc and q2 +50 µc are separated by a distance of d = 12 cm. the electric field at the point p is zero. how far from q1 is p?

Let x be the distance between P and Q1. At point P, the field due to q1 ie. E1 and q2 ie. E2 are equal in magnitude but opposite in direction and hence cancel each other resulting in zero net electric field.

At P

|E1|=|E2|

kq1/x^2 = kq2/(x+d)^2

q1 (x^2+0.0169+0.26x)= q2 × x^2

27x^2 + 0.4563 + 7.02x = 50x^2

23x^2 -7.02x -0.4563 =0

x = 0.36m = 36cm

Or x= -0.055m= -5.5 cm

Note: -5.5 cm corresponds to a position of P between q1 and q2 where the net electric field is also 0.

AFOKE88 AFOKE88 · Answer 2 · 2021-09-09T16:20:50+02:00

This question is dealing with magnitude of electric field.

q1 is 28.97 cm from point P

The equation for the magnitude of an electric field is given as;

E = [tex]\frac{kq}{r^{2}}[/tex],

where;

k is a constant with a value of 8.99 x 10⁹ N.m²/C²

q is magnitude of charge

r is distance from a point

Now, we are told that the electric field at point P is zero.

This means that E_p = 0

Thus; E1 = E2

Image of the charges, distance and the point p are missing and so i have attached it.

Distance of charge q1 from P is x

Distance of charge q2 from P is (x + 12)

q1 = -25 µc (but we will use the absolute value which is 25 µc)

q2 = +50 µc

Thus, using the electric field formula;

[tex]\frac{25k}{x^{2}}[/tex] = [tex]\frac{50k}{(12 + x)^{2} }[/tex]

Simplifying this, k will cancel out and we have;

2[tex]x^{2}[/tex] =

[tex]x^{2}[/tex] - 24 - 144 = 0

Using quadratic formula. we have; x = 28.97 cm

Thus q1 is 28.97 cm from point P

Read more at; brainly.com/question/24245931