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?

Solution:  

Q1 = -27 × 10^-6 C  

Q2 = 50 × 10^-6 C  

d = 13cm = 0.13 m  

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.

the answer is b, have negligible mass

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.