7.1 State Coulomb's Law in words - NSC Physical Sciences - Question 7 - 2017 - Paper 1

Using Coulomb's Law, we have:

$F = k \frac{|q_1 q_2|}{r^2}$

Where:

• $k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2$
• $q_1 = 3 \text{ µC} = 3 \times 10^{-6} \text{ C}$
• $q_2 = 2 \text{ µC} = 2 \times 10^{-6} \text{ C}$
• $r = 0.20 \text{ m}$

Calculating the force:

$F_{JL} = 8.99 \times 10^9 \frac{(3 \times 10^{-6})(2 \times 10^{-6})}{(0.20)^2}$

This simplifies to:

$F_{JL} = 1.35 \text{ N}$.

After contact, the total charge is conserved. The charge on sphere L is $+2 \text{ µC}$ and sphere M initially has $-8 \text{ µC}$. The total charge before contact is:

$Q_{total} = 2 \text{ µC} + (-8 \text{ µC}) = -6 \text{ µC}$

After contact, sphere L and sphere M will share the charge equally since they are conductors. After they separate:

$Q_M = Q_{total} + Q_L = -6 \text{ µC} + 2 \text{ µC} = -4 \text{ µC}$

The amount of charge transferred (Q) can be calculated using:

$Q = ne$

Where

• $e = 1.6 \times 10^{-19} \text{ C is the charge of an electron}.$
  To find $n$ (the number of electrons), rearranging the formula gives:

$n = \frac{Q}{e}$

Assuming sphere L lost charge:

$Q = 2 \text{ µC} = 2 \times 10^{-6} \text{ C}$

Thus:

$n = \frac{2 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 1.25 \times 10^{13}$.

Therefore, approximately $1.25 \times 10^{13}$ electrons were transferred.

To illustrate the electric field pattern:

• The electric field lines for sphere J (positive charge) will radiate outward.
• The field lines for sphere L (after gaining electrons and becoming positively charged) will exhibit similar behavior.
• A sketch should show field lines that start at sphere J and end at sphere L, with no field lines crossing each other.

The net electric field strength ($E_{net}$) at sphere L is the vector sum of the electric field due to sphere J and sphere M:

$E_J = k \frac{|q_J|}{r^2} \ (due\ to\ J)$ $E_M = k \frac{|q_M|}{(r_d)^2} \ (due\ to\ M)$

Where:

• $q_J = +3 \text{ µC}$ (at distance 0.2 m)
• $q_M = -4 \text{ µC}$ (at distance 0.04 m from L after contact)

Plugging in values gives:

$E_J + E_M = \ldots$

Calculating these will give the net electric field strength on sphere L.