If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
1 answer:
Explanation:
The electric field of an isolated charged parallel-plate capacitor is given by :
........(1)
Where
q is the electric charge
A is the area of cross section of parallel plate
It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.
So, the correct option is (E) i.e. "none of the above".
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The question is incomplete. The complete question is :
To measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force F⃗ d on the head of the femur into the hip socket. Then, a horizontal force F⃗ h is applied and increased in magnitude until the femur head rotates clockwise in the socket. The joint is mounted in such a way that F⃗ h will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation.
Draw vectors indicating the normal force n⃗ (magnitude and direction) and the frictional force f⃗ f (direction only) acting on the femur head at point A.
Assume that the weight of the femur is negligible compared to the applied downward force.
Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded
Solution :
The normal force represented by N is equal to the downward force, which is equal in magnitude but it is opposite in direction.
Also the frictional force acts always to oppose the motion because the bone starts moving in a clockwise direction. The frictional force that will be applied to the right direction so that the movement or the rotation at A is opposed.
