Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.?
Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.
(a) What is the electric potential at point a due to q1 and q2?
(b) What is the electric potential at point b?
(c) A point charge q3 = -6.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answer:
a) the potential is zero at the center .
Explanation:
a) since the two equal-magnitude and oppositely charged particles are equidistant
b)(b) Electric potential at point b, v = Σ kQ/r
r = 5cm = 0.05m
k = 8.99*10^9 N·m²/C²
Q = -2 microcoulomb
v= (8.99*10^9) * (2*10^-6) * (1/√2m - 1) / 0.0500m
v = -105 324 V
c)workdone = charge * potential
work = -6.00µC * -105324V
work = 0.632 J
Answer:
1.2 A
Explanation:
From the diagram attached, The three resistors are parallel because the each ends of the resistors are connected together. Since they are in parallel, the voltage across each resistor is the same. The voltage source connected in parallel to the resistors is 60 V. Therefore the voltage across the 50 Ω resistor is 60 V. Using ohm law:
Voltage (V) = Current (I) × Resistance (R)
V = IR
I = V/R
I = 60 V/ 50 Ω
I = 1.2 A
The current in the 50 Ω resistor is 1.2 A
Answer:
An object has potential energy (stored energy) when it is not in motion. Once a force has been applied or it begins to move the potential energy changes to kinetic energy (energy of motion).
EXAMPLE: A rock sitting on the edge of a cliff. If the rock falls, the potential energy will be converted to kinetic energy, as the rock will be moving. A stretched elastic string in a longbow.
the answer is A. The aluminum has 0.84 ohms more resistance.
Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
