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(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?

Short Answer

Expert verified

The charge resides on the metal ball is \( - 5.11 \times {10^{ - 11}}\;{\rm{C}}\).

Step by step solution

01

Understanding the electric field

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 \(9.0 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\).

02

Given information:

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

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

03

Determination of the charge that resides on the ball

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 \( - 5.11 \times {10^{ - 11}}\;{\rm{C}}\).

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Most popular questions from this chapter

Refer to Fig. 16โ€“32d. If the two charged plates were moved until they are half the distance shown without changing the charge on the plates, the electric field near the center of the plates would

(a) remain almost exactly the same.

(b) increase by a factor of 2.

(c) increase, but not by a factor of 2.

(d) decrease by a factor of 2.

(e) decrease, but not by a factor of 2.

A point charge of mass 0.185 kg, and net charge\( + {\bf{0}}{\bf{.340 }}\mu {\bf{C}}\)hangs at rest at the end of an insulating cord above a large sheet of charge. The horizontal sheet of fixed uniform charge creates a uniform vertical electric field in the vicinity of the point charge. The tension in the cord is measured to be 5.18 N. Calculate the magnitude and direction of the electric field due to the sheet of charge (Fig. 16โ€“67).

FIGURE 16โ€“67 Problem 61.

Measurements indicate that there is an electric field surrounding the Earth. Its magnitude is about 150 N/C at the Earthโ€™s surface and points inward toward the Earthโ€™s center. What is the magnitude of the electric charge on the Earth? Is it positive or negative? (Hint: The electric field outside a uniformly charged sphere is the same as if all the charge were concentrated at its center.)

(I) What is the magnitude of the force a\({\bf{ + 25}}\;{\bf{\mu C}}\)charge exerts on a\({\bf{ + 2}}{\bf{.5}}\;{\bf{mC}}\)charge 16 cm away?

A 3.0-g copper penny has a net positive charge of \({\bf{32}}\;{\bf{\mu C}}\). What fraction of its electrons has it lost?

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