<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency 
Velocity of the string wave is equal to 
Power of wave propagation is equal to 
So power of the wave will be equal to 5.464 watt
the first step would be gathering information , and making sure its correct
Answer:
Explanation:
We have to find electric potential V at a distance r.
a) For r>R,
The electric field in the cylinder is given by
E.A equating it to the other electric field given by
б.A/ε₀
Here the area of cylinder is given by= 2*3.14*r*L
While for the outside, the area= 2*3.14*R*L
Equating both, we get
E= бR/rε₀
Now,
The potential difference is given as:
ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.
Where ΔV= V₀-V
So solving we get
V₀=V-бR/ε₀ln (r/R)
b) For r<R i.e. inside the cylinder
There will be no electric field produced as E=0
So ultimately Vin= V
c) V=0 at r= infinity.
Heres your great lovley answer that you need
