<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
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energy is the correct answer to fill the blank bb :)
D multiply force
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Answer:
Net Force = 10N
Acceleration = 2m/s^2
Explanation:
calculate the net force and the acceleration on the block
Net force on the block F = mass * acceleration
Net force acting in the positive direction = 4N + 6N = 10N
Mass = 5kg
According to newton's second law;
a = F/m
a = 10N/5
a = 2m/s^2
hence the acceleration on the block is 2m/s^2
