<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:
The last statement is false.
Explanation:
Photons (Electromagnetic radiation) are released when electrons drop from a higher energy lever to a lower energy level. Therefore the opposite insinuated by the last statement is wrong.
Answer:
5.71428571422 m/s
Explanation:
u = Initial velocity = 20 m/s
v = Final velocity
s = Displacement
a = Acceleration
Time taken = 15-1 = 14 s
Distance traveled in 1 second = 
The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s
Answer:
As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.
Formula: g = GM/r2
Dimensional Formula: M0L1T-2
Values of g in SI: 9.806 ms-2
Explanation:
Please Mark me brainliest
An external force that is being applied in the direction of the displacement
