Question

Identical 11.8 mC charges are fi xed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fi xed to one of the empty corners, so that the total electric potential at the remaining empty corner is 0 V

Answer 1

Approximately
`(-20.1)\; {\rm mC}`.

Step-by-step explanation:

By Coulomb's Law, at a distance of
`r` from an electric charge
`q`, the electric potential resulting from that charge would be:

`\displaystyle E = (k\, q)/(r)`,

Where
`k` is Coulomb's Constant.

When there are more than one electric charges nearby, the resultant electric potential is the scalar sum of the potential from each of these charges.

Let
`q_(1) = 11.8\; {\rm mC}` denote the two charges at adjacent vertices of this square. Let
`q` denote the unknown charge.

Let
`r` denote the length of each side of this square. At the empty vertex:

• Distance from one of the
  `q_(1) = 11.8\; {\rm mC}` charge would be
  `r`;
• Distance from the other
  `q_(1) = 11.8\; {\rm mC}` charge would be
  `\left(√(2)\right)\, r` (along the diagonal of the square.)
• Distance from the unknown charge
  `q` would be
  `r`.

The resultant electric potential at the empty vertex would be the sum of the potential from each of these charges:

`\displaystyle (k\, q_(1))/(r) + (k\, q_(1))/(\left(√(2)\right)r) + (k\, q)/(r)`.

Set this expression value to
`0` and solve for
`q`.

`\displaystyle (k\, q_(1))/(r) + (k\, q_(1))/(\left(√(2)\right)r) + (k\, q)/(r) = 0`.

Note that
`k` and
`r` are both non-zero constants and can be eliminated. Hence:

`\displaystyle q_(1) + (q_(1))/(√(2)) + q = 0`.

`\begin{aligned} q &= -\left(q_(1) + (q_(1))/(√(2))\right) \\ &= -\left((2 + √(2))/(2)\right)\, q_(1) \\ &= -\left((2 + √(2))/(2)\right)\, (11.8\; {\rm mC}) \\ &\approx (-20.1)\; {\rm mC}\end{aligned}`.

In other words, the unknown charge should be
`(-20.1)\; {\rm mC}` to ensure that the electric potential is zero at the empty vertex.