The electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
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Electric field of the positive particle</h3>
The electric field is calculated as follows;
E = kq/r²
where;
- r is the distance between the charges
- k is Coulomb's constant
- q is magnitude of the charge
midpoint of 3.08 m, x = 1.54 mm
r(1.54 mm, 2.00 mm)
|r| = √(1.54² + 2²)
|r| = 2.52 mm
E = (9 x 10⁹ x 37.3 x 10⁻³)/(2.52 x 10⁻³)²
E = 5.287 X 10¹³ N/C
Thus, the electric field at point A located 2.00 mm above the dipole's midpoint is 5.287 X 10¹³ N/C.
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Answer:
0.5 A
Explanation:
Applying,
V = IR.................. Equation 1
Where V = Voltage, I = current, R = Resistance.
make I the subject of the equation
I = V/R............... Equation 2
From the question,
Given: V = 75 V, R = 150 Ω
Substitute these values into equation 2
I = 75/150
I = 0.5 A.
Hence the cuurent through the resistor is 0.5 A
I can why do plants grow fast?
Inertia is that quantity which depends solely upon mass. The more mass, the more inertia. Momentum is another quantity in Physics which depends on both mass and speed.
