At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi
1 answer:
Answer:
The distance from the charge is 3.35 m.
Explanation:
Given that,
Electric potential, V = 635 V
Magnitude of electric field, E = 189 N/C
We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :
d is the distance from charge
So, the distance from the charge is 3.35 m. Hence, this is the required solution.
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<h2>Answer:</h2>
1.8 x 10⁻⁵J
<h2>Explanation:</h2>The energy (E) stored in a capacitor of capacitance, C, when a voltage, V, is supplied is given by;
E = x C x V² -------------------(i)
Now, from the question;
C = 2.00μF = 2.00 x 10⁻⁶F
V = 18.0V
Substitute these values into equation (i) as follows;
E = x 2.00 x 10⁻⁶ x 18.0
E = 1.8 x 10⁻⁵J
Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J