A charge of 1.5 × 1015 coulombs moves from point A to a lower potential at point B. The electric field is uniform at 2.5 × 10-5
1 answer:
The work done by the electric field is given by:
where q is the charge, and
is the potential difference between point A and B. But the potential difference is equal to
where E is the intensity of the electric field, while d is the distance between the two points A and B.
Therefore, the work done by the field is
where q is the charge, and
where E is the intensity of the electric field, while d is the distance between the two points A and B.
Therefore, the work done by the field is
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Answer:
9.81 × 10⁻¹⁰ C
Explanation:
Given:
Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m
Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C
Mass of the tissue = 3 g = 0.003 Kg
Now,
The force required to pick the tissue, F = mg
where, g is the acceleration due to gravity
also,
The force between (F) the charges is given as:
where,
q is the charge on the tissue
k is the Coulomb's constant = 9 × 10⁹ Nm²/C²
thus,
on substituting the respective values, we get
or
q = 9.81 × 10⁻¹⁰ C
Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C
Answer:
Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.
Explanation:
Considering the complete question attached in figure below.
Time period for balance wheel is:
m = mass of balance wheel
R = radius of balance wheel.
Angular frequency is related to Time period as:
As dimensions of new balance wheel are one-third of their original values
