Point charges of 5.00 µC and − 3.00 µC are placed 0.250 m apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?

a. 0.8620m

b. 0.1092m

Explanation:

a.

Given

Q1 = 5.00 μC and

Q2 = -3.00 μC

Let Q3 = the third charge (the net force on Q3 = 0)

Let d = distance between Q1 and Q2 = 0.250m

Assume Q3 is placed at a distance x to the right of Q2.

Then d + x = the distance between the Q3 and Q1.

The distance the third charge be placed so that the net force on it is zero is given as.

k|Q1||Q3|/(d+x)² = k|Q2||Q3|/(x)²

Make x the subject of formula

x = d/±√((|Q1|/|Q2|) - 1)

Substitute each values

x = 0.250m/±√((|5|/|-3|) - 1)

x = 0.250/±√((5/3) - 1

x = 0.250/(±1.290 - 1)

x = 0.250/(1.290 - 1) or 0.250/(-1.290 - 1)

x = 0.8620 or −0.1092

But x can't be negative.

So, x = 0.8620

b.

Q1 = 5.00 μC and

Q2 = -3.00 μC

Let Q3 = the third charge (the net force on Q3 = 0)

Let d = distance between Q1 and Q2 = 0.250m

Assume Q3 is placed at a distance x to the right of Q2.

Then d - x = the distance between the Q3 and Q1.

The distance the third charge be placed so that the net force on it is zero is given as.

k|Q1||Q3|/(d-x)² = k|Q2||Q3|/(x)²

Make x the subject of formula

x = d/(1±√((|Q1|/|Q2|))

Substitute each values

x = 0.250m/(1±√((|5|/|3|))

x = 0.250/(1±√((5/3))

x = 0.250/(1±1.290)

x = 0.250/(1 + 1.290) or 0.250/(1 - 1.290)

x = 0.1092 or −0.8620

But x can't be negative.

So, x = 0.1092

answer:

an example of solubility is salt or sugar dissolving in water.

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