Answer:
Answer in explanation.
Explanation:
The relationship between the charge on the capacitor and the potential difference across it is given as follows:
where,
Q = Charge on the Capacitor
C = Capacitance of the Capacitor
V = Potential Difference across the Capacitor
This relationship can be used to find the charge on a capacitor, using the voltmeter, as follows:
<u>The potential difference can be measured through the voltmeter. And the capacitance of the capacitor is a known constant value. Therefore, the charge can be found by taking product of both.</u>
Answer:
(a) Electrostatic force F=4.51×10⁻¹⁷N
(b) Number of electron n=4425 electrons
Explanation:
Given data
Charges q₁=q₂= -7.08×10⁻¹⁶C
Distance r=1.00cm =0.01 m
To find
(a) Electrostatic force F
(b) Number of electron n
Solution
For (a) Electrostatic force
From Coulombs law we know that
For (b) number of electron n
The number of electron on the drop that giving it its imbalance is the total charge divided by charge of electron
As we now that charge of electron e= -1.6×10⁻¹⁹C
Answer:
(a)
(b)
Explanation:
Let us take the north direction to be the positive y-axis and the east to be positive x-axis.
First day:
25.0 km southeast, which implies south of east. The y-component will be negative and the x-component will be positive.
Second day:
She starts off at the stopping point of last day. This time, both the y- and x-components are positive.
Therefore, total displacements:
Magnitude of displacements,
Direction,