derive an expression for electric potential due to point charge
Answers
Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q, and noting the connection between work and potential \left(W=\phantom{\rule{0.25em}{0ex}}-q\Delta V\right), it can be shown that the electric potential V of a point charge is
V=\frac{\text{kQ}}{r}\phantom{\rule{0.25em}{0ex}}\left(\text{Point Charge}\right),
where k is a constant equal to
9.0×{\text{10}}^{\text{9}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}\text{·}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{\text{2}}\text{/}{\text{C}}^{\text{2}}.
Electric Potential V of a Point Charge
The electric potential V of a point charge is given by
V=\frac{\text{kQ}}{r}\phantom{\rule{0.25em}{0ex}}\left(\text{Point Charge}\right).
The potential at infinity is chosen to be zero. Thus V for a point charge decreases with distance, whereas \mathbf{\text{E}} for a point charge decreases with distance squared:
\text{E}=\frac{\text{F}}{q}=\frac{\text{kQ}}{{r}^{2}}.
Recall that the electric potential V is a scalar and has no direction, whereas the electric field \mathbf{\text{E}} is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas \mathbf{\text{E}} is closely associated with force, a vector.
⭕⭕⭕❤❤❤Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q, and noting the connection between work and potential \left(W=\phantom{\rule{0.25em}{0ex}}-q\Delta V\right), it can be shown that the electric potential V of a point charge is
V=\frac{\text{kQ}}{r}\phantom{\rule{0.25em}{0ex}}\left(\text{Point Charge}\right),
where k is a constant equal to
9.0×{\text{10}}^{\text{9}}\phantom{\rule{0.25em}{0ex}}\text{N}\phantom{\rule{0.25em}{0ex}}\text{·}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{\text{2}}\text{/}{\text{C}}^{\text{2}}.
Electric Potential V of a Point Charge
The electric potential V of a point charge is given by
V=\frac{\text{kQ}}{r}\phantom{\rule{0.25em}{0ex}}\left(\text{Point Charge}\right).
The potential at infinity is chosen to be zero. Thus V for a point charge decreases with distance, whereas \mathbf{\text{E}} for a point charge decreases with distance squared:
\text{E}=\frac{\text{F}}{q}=\frac{\text{kQ}}{{r}^{2}}.
Recall that the electric potential V is a scalar and has no direction, whereas the electric field \mathbf{\text{E}} is a vector. To find the voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a scalar, whereas \mathbf{\text{E}} is closely associated with force, a vector.
Hope it helps uh❤❤❤❤
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