A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.
<h3>What is electric potential energy?</h3>
The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.
The electric potential energy is given by the relation,
V = kQ/r
where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.
Substitute the values into the expression to get the distance between the charges.
8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r
r =3.34 x 10⁷ m
Thus, the distance between the charges will be 3.34 x 10⁷ m.
Learn more about electric potential energy.
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-- The object either left or crossed the starting line exactly at time=0 .
-- The object has been traveling at constant speed for all time that
we know about.
Answer:
the Answer Would be D
Explanation:
From what I know light travels 300,000 km/second , travels at fast speeds and travels in a straight line
Answer:
(a) B = 2.85 ×
Tesla
(b) I = I = 0.285 A
Explanation:
a. The strength of magnetic field, B, in a solenoid is determined by;
r = 
⇒ B = 
Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field
B = 
= 
B = 2.85 ×
Tesla
b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;
B = μ I N/L
⇒ I = B ÷ μN/L
where B is the magnetic field, μ is the permeability of free space = 4.0 ×
Tm/A, N/L is the number of turns per length.
I = B ÷ μN/L
= 
I = 0.285 A
Answer:
240 Ω
Explanation:
Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).
The expression for resistance power and voltage is give as,
P = V²/R.......................... Equation 1
Where P = Power, V = Voltage, R = Resistance
Making R the subject of the equation,
R = V²/P.................... Equation 2
Given: V = 120 V, P = 60 W.
Substitute into equation 2
R = 120²/60
R = 240 Ω
Hence the resistance of the bulb = 240 Ω