Sir solve it
Asked by venkat
24th October 2018,
3:58 PM
Answered by Expert
Answer:
Let us assume the X-axis along the direction from centre of the clock to the number 3 on the dial as shown in figure.
Similarly Y-axis along the direction from centre to number 12 in the dial.
Let us assume a unit +ve test charge is placed at centre. Along the line connecting the number 1 and number 7,
there is a net force of 6q/(4πε0r2 ) from centre to number 7, where r is radial distance from centre to numbers
and ε0 is permitivity of free space. If we resolve this net force along X and Y axis, we get the respective components
as -3qK and -3√3qK, where K = 1/(4πε0r2 ).
Table given below shows the resolved components of net force acting along the lines joining the other numbers.
(angles are measured from X-axis. Anti-clockwise direction +ve, clockwise direction -ve)
After getting all the resolved components we add them to get the resultant force.
Right side of the figure given above shows the net force acting along X an Y direction and the resultant is found to
make an angle 15º with the -ve X-axis direction. This direction is in between number 9 and 10.
Hence when the time is 9-30, hour hand will be along this direction of resultant force or field direction.
|
Line of Action Of force |
X-component (N) |
Y-component (N) |
|
‘1’ and ‘7’ |
6qK cos(-120) = -3qK |
6qK sin(-120) = -3√3qK |
|
‘2’ and ‘8’ |
6qK cos(-150) = -3√3qK |
6qK sin(-150) = -3qK |
|
‘3’ and ‘9’ |
-6qK |
0 |
|
‘4’ and ‘10’ |
6qK cos(150) = -3√3qK |
6qK sin(150) = 3qK |
|
‘5’ and ‘11’ |
6qK cos(120) = -3K |
6qK sin(120) = 3√3qK |
|
‘6’ and ‘12’ |
0 |
6qK |
|
|
Total = (-12-6√3)qK |
Total = 6qK |
Answered by Expert
25th October 2018,
3:28 AM
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