In BPC
tan\theta =a/b = 3/4
\theta = tan^-1(0.75)
\theta = 36.87 deg
BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m
Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C
Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C
Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C
Net electric field along X-direction is given as
Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C
Net electric field along X-direction is given as
Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C
Net electric field is given as
E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
Answer:
Explanation:
(a) For the calculation of the Electric field we use

(b) The capacitance is calculate by using the expression

(c) Finally, the charge on each plate is

I hope this is useful for you
Regards
It's because lightbulbs need to carry an electric current within a range of resistance so that the material doesn't without breaking down.
Answer:
They collide, couple together, and roll away in the direction that <u>the 2m/s car was rolling in.</u>
Explanation:
We should start off with stating that the conservation of momentum is used here.
Momentum = mass * speed
Since, mass of both freight cars is the same, the speed determines which has more momentum.
Thus, the momentum of the 2 m/s freight car is twice that of the 1 m/s freight car.
The final speed is calculated as below:
mass * (velocity of first freight car) + mass * (velocity of second freight car) = (mass of both freight cars) * final velocity
(m * V1) + (m * V2) = (2m * V)
Let's substitute the velocities 1m/s for the first car, and - 2m/s for the second. (since the second is opposite in direction)
We get:

solving this we get:
V = - 0.5 m/s
Thus we can see that both cars will roll away in the direction that the 2 m/s car was going in. (because of the negative sign in the answer)