Answer:
-30 N/C
Explanation:
Since the potential changes from 0.90 V to 1.2 V when I move the probe 1 cm closer to the non-grounded electrode, the electric field is the gradient between the two points is given by E = -ΔV/Δx where ΔV = change in electric potential and Δx = distance of potential change = 1 cm = 0.01 m
Now ΔV = final potential - initial potential = 1.2 V - 0.90 V = 0.30 V
Since E = -ΔV/Δx
substituting the values of the variables into the equation, we have
E = -ΔV/Δx
E = -0.30 V/0.01 m
E = -30 V/m
Since 1 V/m = 1 N/C.
E = -30 N/C
So, the average electric field is -30 N/C
It appears to be a <span>spiral shape. </span>
- m1=1500kg
- m_2=3000kg
- v_1=5m/s
- v_2=7m/s
Using law of conservation of momentum
When the object is big enough to contract itself into a ball.
Answer:
The value of each charge is 4.22 x 10⁻⁵ C
Explanation:
Given;
distance between the two identical charges, d = 2 m
the force of repulsion between these two charges, F = 4N
Apply Coulomb's law;
Therefore, the value of each charge is 4.22 x 10⁻⁵ C
