it's far the amount of work energy that is needed to circulate a charge towards the electric field.
electric capability because of a unmarried fee: \(V = k \dfrac{q}{r}\)
Where;
remember we've got extraordinary point fees that are inflicting the electrical capacity! Now right here, each rate places stress at the point fee which could without difficulty be calculated thru this electric powered potential calculator online. Mathematically, we've got the subsequent electric powered ability energy formulas for each price:
\(V_1 = k \dfrac{q_1}{r_1}\)
\(V_2 = k \dfrac{q_2}{r_2}\)
\(V_3 = k \dfrac{q_3}{r_3}\)
\(V_4 = k \dfrac{q_4}{r_4}\)
here, the electric ability electricity calculator uses the superposition principle to calculate the electrical capability difference on a median.
\(V = V_1+V_2+V_3+V_4\)
\(V = k \left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}+\dfrac{q_3}{r_3}+\dfrac{q_4}{r_4}\right)\)
\(V = V_1+V_2+V_3+V_4+...+V_n\)
\(V = k \left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}+\dfrac{q_3}{r_3}+\dfrac{q_4}{r_4}+...+\dfrac{q_n}{r_n}\right)\)
If we are given a price of (four*10^12C) and a distance of about 2 cm, how to find electric capacity?
the use of the electric capacity system:
\(V = k \dfrac{q}{r}\)
\(V = \dfrac{1}{4πε_{o}}*\dfrac{4*10^12C}{2 cm}\)
\(V = \dfrac{1}{4*3.14*1} \dfrac{4000000000000}{0.002m}\)
\(V = 1.798*10^+24\)
that's the specified potential. To affirm it, virtually input the charge and distance in the electrostatic ability energy calculator and it's going to output the consequences. After that, in shape the result with the calculated right here.
actually sure! Recalling the announcement that electrical potential is proportional to the nature of the rate, a bad charge will cause a bad electric powered capacity.
As the electrical capacity is proportional to the distance between the chargers, the capacity at a factor at infinity may be zero.
1 electric powered ability is described as the electric capacity power according to unit fee..
