Question

Two fixed particles, of charges qı = +3.0 mC and qı = +1.0 mC, are 10 cm apart. How far from q, should a third charge be located at distance x so that no net electrostatic force acts on it? X q1 43 92 a) 3.66 cm, b) 4.49 cm, c) 6.34 cm d) 4.14 cm.

Answer 1

Approximately
`6.34\; {\rm cm}` from the
`(+3.0)\; {\rm mC}` charge, which is approximately
`10\; {\rm cm} -6.34\; {\rm cm} \approx 3.66\; {\rm cm}` from the
`(+1.0)\; {\rm mC}` charge.

Step-by-step explanation:

To find the position of the charge, start by determining the position of this new charge relative to the two fixed charges- between the two charges, or on the same side of the two charges. After that, apply Coulomb's Law to find the magnitude of the force from each fixed charge. Set the two magnitudes to be equal and solve the equation to find the position of the required charge.

Forces from the two fixed charge would be balanced only if the two forces are in opposite directions. Assuming that the new charge is positive, it will repel both fixed charges:

• If the new charge is to the left of both fixed charges, forces from both fixed charge would point to the left (further away from the fixed charges.) Hence, the resultant force would be nonzero.
• If the new charge is to the right of both fixed charges, forces from both fixed charges would point to the right. The resultant force would also be nonzero.
• If the new charge is between the two fixed charges, the force from the fixed charge on the left would point to the right, while the force from the fixed charge on the right would point to the left. The resultant force could be balanced since the two forces point in opposite directions.

Similarly, it can be shown that if the new charge is negative, the force from the two fixed charges would be balanced only if the new charge is between the two fixed charges.

Assume that the new charge is between the two fixed charges and is at a distance of
`x\; {\rm cm}` from the
`(+3.0)\; {\rm mC}` charge (
`0 < x < 10`.)

Since the
`(+3.0)\; {\rm mC}\!` fixed charge is at a distance of
`10\; {\rm cm}` from the
`(+1.0\; {\rm mC})` fixed charge, the distance between the new charge and the
`(+1.0\; {\rm mC})\!` fixed charge would be
`(10 - x)\; {\rm cm}`.

By Coulomb's Law, when two charges of magnitude
`q_(1)` and
`q_(2)` are at a distance of
`r` from each other, the magnitude of the electrostatic force between them would be:

`\displaystyle (k\, q_(1)\, q_(2))/(r^(2))`,

Where
`k` is the coulomb constant.

Let
`q` denote the magnitude of the new charge. Magnitude of the force between the new charge and the
`(+3.0)\; {\rm mC}` charge would be:

`\displaystyle \frac{k\, (q)\, (3.0\; {\rm mC})}{(x\; {\rm cm})^(2)}`.

Magnitude of the force between the new charge and the
`(+1.0\; {\rm mC})` charge would be:

`\displaystyle \frac{k\, (q)\, (1.0\; {\rm mC})}{((10 - x)\; {\rm cm})^(2)}`.

Equate the two expressions and solve for
`x`:

`\displaystyle \frac{k\, (q)\, (3.0\; {\rm mC})}{(x\; {\rm cm})^(2)} = \frac{k\, (q)\, (1.0\; {\rm mC})}{((10 - x)\; {\rm cm})^(2)}`.

`\displaystyle (3.0)/(x^(2)) = (1.0)/((10 - x)^(2))`.

`x^(2) - 3.0\, (10 - x)^(2) = 0`.

The two real roots of this quadratic equation are
`x \approx 23.7` and
`x \approx 6.34`. Since the new charge needs to be located between the two fixed charges,
`x \approx 6.34\!` is the only valid solution.

Therefore, the new charge should be approximately
`6.34\; {\rm cm}` from the
`(+3.0)\; {\rm mC}` charge, which is approximately
`10\; {\rm cm} - 6.34\; {\rm cm} \approx 3.66\; {\rm cm}` from the
`(+1.0)\; {\rm mC}` charge.