Q2: Find the magnitude of the magnetic force acting on the ball just before it h

Question

Q2: Find the magnitude of the magnetic force acting on the ball just before it hits the ground

The rod is maintained at a potential of -5000 volts with respect to the square which is grounded (V=0). The contour lines are equipotentials with V = -4000, -3000, -2000 and -1000 volts. Select a response for each statement below (Use 'N', North, towards top of page (+y dir), and 'E, East, to the right (+x dir), 'NE' for in between 'N' and 'E', etc.) The direction of the force on a small (+) charge at 2 is .... The magnitude of the E-field at 5 is greater than that at 4 The direction of the force on a small (-) charge at 8 is .... The x-component of the E-field at 2 is zero. The x-component of the E-field at 4 is negative. The magnitude of the E-field at 6 is very small (~ zero). The direction of the Efield at 7 is .... Q2: Find the magnitude force acting on the ball just before it hits the ground

Explanation / Answer

a)

We know that +ve charges move from a region of higher electric potential to lower potential. From the figure we see that the lowest potential(-5000V) is at the rod and subsequently the potential increases(-5000 V increases to -1000 V ---- NOTE---> -ve value decreases so potential increases) with increase in distance fro the rod.

So, the potential will be highest at far away from the rod.

So, for a charge at 2, it will tend to move towards the rod(as +ve charge will try to move towards decreasing potential region). So force will be towards inner circle(ie towards N )..

b)

E-field is the rate of change of potential. So the dencity of equipotential lines gives the E-field.

At 5, the lines are dense so E-field at 5 is greater than at 4....

c)

For -ve charges, the effect is opposite. So force will be outwards. So force will be towards North(N)

d)

As the equipotential lines at 2 are parallel. So E-field lines must be perpendicular to 2. So, the x-component of E-field is 0 <-------answer

e)

x-component of E-field at 4 must have +ve value

f)

at far away places(like at point 6) the electric equipotnetial lines will be very less dense.So E-field will be very less

g)

dircetion will be N-E(north east)

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