Point P is located 58 cm to the right of a fixed point charge of 6.7*10^ -16 C. What is the direction and magnitude of the elect
1 answer:
The magnitude of the electric field is , the direction is to the right
Explanation:
The magnitude of the electric field produced by a single point charge is given by
where:
is the Coulomb's constant
is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
is the charge
r = 58 cm = 0.58 m is the distance of point P from the charge
Substituting, we find the magnitude of the field at point P:
The direction of the field for a positive charge is away from the charge: therefore, since point P is to the right of the charge, the direction of the field is to the right.
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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
(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
<h3>Total capacitance of the circuit</h3>
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
<h3>C1 and C2 are in series </h3>Ct + C(78) = 2 μF + 3 μF = 5 μF
<h3 /><h3>Total charge stored in the circuit</h3>The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
<h3>Charge stored in 3μF capacitor</h3>Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
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