Complete Question:
Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
A. 2F
B. F/4
C. F/2
D. F
E. 4F
Answer:
D.
Explanation:
If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.
As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.
Answer:
0.1835m/s
Explanation:
The formula for calculating the speed of wave is expressed as;
v = fλ
f is the frequency - The number of oscillations completed in one seconds
If 22 waves pass the boat every 60 seconds,
number of wave that passes in 1 seconds = 22/60 = 0.367 waves
Therefore the frequency f of the wave is 0.367Hertz
λ (wavelength) is the distance between successive crest and trough of a wave
λ = 0.5m
Substitute the given values into the formula
v = fλ
v = 0.367 * 0.5
v = 0.1835
Hence the speed of the waves is 0.1835m/s
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Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:

Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):

(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):

(towards)
Now, the total field at the origin is the sum of both the fields:
