If object is not accelerating, the sum of all forces on the object will be equal to ZERO...
<span>Radius, the distance from the centre = 0.390
Electric field is equal to half of the magnitude. E2 = E / 2
Given
E1 = E2
E1 = k x Q / r^2
E2 = (k x Q / r2^2) / 2
Equating the both we get 2 x r^2 = r2^2
r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
r2 = 1.414 x 0.390 = 0.551 m</span>
The answer will be C based on what I had researched
Answer:
As per Coulomb's law we know that force between two charges is given as
F = \frac{kq_1q_2}{r^2}F=
r
2
kq
1
q
2
here we know that
q_1 = 2.5 \times 10^{-6} Cq
1
=2.5×10
−6
C
q_2 = -5.0 \times 10^{-6} Cq
2
=−5.0×10
−6
C
r = 0.0050 mr=0.0050m
now from above formula we will have
F = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(5 \times 10^{-6})}{(0.0050)^2}F=
(0.0050)
2
(9×10
9
)(2.5×10
−6
)(5×10
−6
)
F = 4500 NF=4500N
so they will attract towards each other as they are opposite in nature with force F = 4500 N
