Answer :
123.61KV difference is the electric potential 20 cm from a charge of 5.5 c compared to 40 cm from the same charge.
electric force F=[tex]\frac{KQq}{r^2}[/tex]
electric potential V=[tex]\frac{Fr}{Q}[/tex]
V=[tex]\frac{Fq}{r}[/tex]
this equation we know that [tex]V\propto \frac{1}{r}[/tex]
the point closer to the charge is higher than the point farther away.
B) for 20cm
V=[tex]\frac{kq}{r}[/tex]
V=[tex]\frac{8.99*10^9*5.5*10^{-6}}{0.2}[/tex]
V=247.25KV
for 40cm
V=[tex]\frac{8.99*10^9*5.5*10^{-6}}{0.4}[/tex]
V=123.61KV
∴ the difference is 247.25-123.61= 123.61KV
the electric capability (additionally known as the electric discipline capability, potential drop, the electrostatic capacity) is defined as the amount of work strength had to move a unit of electric fee from a reference point to the specific factor in an electric-powered field.
the electric-powered potential is the work done consistent with a unit fee to deliver the rate from infinity to a degree in an electric discipline. the capacity distinction is the distinction between the potentials between two factors within the electric-powered discipline. electric powered ability is described as a factor. In an electrical circuit, the capability between factors (E) is defined as the quantity of labor carried out (W) by using an external agent in shifting a unit fee (Q) from one point to another. Mathematically we will say that E = W/Q.electric-powered capability is a scalar, and electric-powered discipline is a vector. The addition of voltages as numbers offers the voltage due to a combination of point costs, whereas the addition of man or woman fields as vectors offers the full electric-powered subject.
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