A small cylinder is rolled along a ruler and completes two revolutions. The circumference is the distance around the outside of
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Answer:
Incomplete question check attachment for complete question
Also it is given that
q1=-0.7uC
q4=-1.7uC
q3=-1.7uC
Also the distance are given as
a=2.2cm=0.022m
d2=3.6cm=0.036m
Explanation:
The potential energy due to point R is given as
The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4
So, let take it one after the other
Potential energy is give as
P.E=kq1q2/r
Therefore,
Potential energy due to charge q1 and q3
U¹³=kq1q3/r
To get the distance between charge q1 and q3, we will apply Pythagoras theorem
r=√(d2²+a²)
r=√(0.036²+0.022²)
r=0.0422m
k is a constant =9×10^9Nm²/C²
Then,
U¹³=kq1q3/r
U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422
U¹³=0.254J
Potential energy due to charge q1 and q4
U¹⁴=kq1q4/r
To get the distance between charge q1 and q4, we will apply Pythagoras theorem
r=√(d2²+a²)
r=√(0.036²+0.022²)
r=0.0422m
k is a constant =9×10^9Nm²/C²
Then,
U¹⁴=kq1q4/r
U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422
U¹⁴=0.254J
Potential energy due to charge q3 and q4
U³⁴=kq3q4/r
r=2a=2×0.022=0.044m
k is a constant =9×10^9Nm²/C²
Then,
U³⁴=kq3q4/r
U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044
U¹⁴=0.591J
Then, the total energy is
U= U¹³+ U¹⁴ + U³⁴
U=0.254+0.254+0.591
U=1.099J
Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.
