<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:
yes This is correct Answer
The frictional force is 39.4 N
Explanation:
We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write
where
is the net force
m is the mass
a is the acceleration
Here we know that the box is moving with constant velocity, so its acceleration is zero:
This means that the net force is also zero:
The net force on the block is given by the applied force, forward, and the frictional force, backward:
where
is the applied force
is the frictional force
Therefore, solving for ,
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Yes that's correct. Also zeros in between non-zero numbers are significant figures
