(II) The field just outside a 3.50-cm-radius metal ball is \({\bf{E = 3}}{\bf{.75 \times 1}}{{\bf{0}}^{\bf{2}}}\;{\bf{N/C}}\) and points toward the ball. What charge resides on the ball?

The value of electric field (E) due to a charge Q at any point is determined by finding the electric force (F) per unit charge acting on a small positive test charge (q) placed at that point.

The expression for electric field is given as:

\(E = \frac{F}{q} = k\frac{Q}{{{r^2}}}\)

Here, k is the electrostatic force constant whose value is \(9.0 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\).

The electric field outside the metal ball is, \(E = 3.75 \times {10^2}\;{\rm{N/C}}\)

The radius of the metal ball is, \(r = 3.50\;{\rm{cm}} = 3.50 \times 1{{\rm{0}}^{ - 2}}\;{\rm{m}}\)

If E is the electric field due to charge placed inside the metal ball at its center, then the magnitude of electric field due to the charge Q placed at a distance r from it is:

\(E = k\frac{Q}{{{r^2}}}\)

So, the charge inside the metal ball is given as:

\(Q = \frac{{E{r^2}}}{k}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}Q = \frac{{\left( {3.75 \times {{10}^2}\;{\rm{N/C}}} \right) \times {{\left( {3.50 \times 1{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)}^2}}}{{\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\ = 5.11 \times {10^{ - 11}}\;{\rm{C}}\end{aligned}\)

Since any charge in a metallic conductor resides on the surface of the conductor, the magnitude of charge that resides on the metal ball is \(5.11 \times {10^{ - 11}}\;{\rm{C}}\). Also, as the electric field points towards the metal ball, the charge on the metal ball must be negative.

Thus, the charge that resides on the surface of the metal ball is \( - 5.11 \times {10^{ - 11}}\;{\rm{C}}\).