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Chapter 23: Q. 17 (page 654)

The electric field strength $2.0 cm$ from the surface of a $10 cm$ diameter metal ball is $50, 000 N / C$ . What is the charge $(in nC)$ on the ball?

Short Answer

Expert verified

Charge on the ball is $27.2 nC$ .

Step by step solution

01

Formula for finding charge

Metal sphere has radius of $5 cm$ .

Electric field at from surface is $50, 000 N / C$ .

Electric field due to sphere of radius $R$ and charge $Q$ at distance is $r \geq R$

$E = \frac{Q}{4 {πϵ}_{o} R^{2}} . . . . . . (1)$

02

Calculation for charge

Using equation $(1)$ ,

Calculate charge $Q$ at metal sphere.

$Q = E (4 {πϵ}_{o} R^{2})$

$= (50, 000 N / C) ({(9 \times 10^{9})}^{- 1} (0.02 m + 0.05 m)^{2})$

localid="1648628325367" $= (50, 000 N / C) ({(9 \times 10^{9})}^{- 1} (0.07)^{2})$

$= (50, 000 N / C) ({(9 \times 10^{9})}^{- 1} (0.0049))$

$Q = 27.2 nC$

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Most popular questions from this chapter

Air "breaks down" when the electric field strength reaches $3.0 \times 10^{6} N / c$ , causing a spark. A parallel-plate capacitor is made from two $4.0 c m \times 4.0 c m$ electrodes. How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?

The surface charge density on an infinite charged plane is $- 2.0 \times 10^{- 6} C / m^{2}$ . A proton is shot straight away from the plane at $2.0 \times 10^{6} m / s$ . How far does the proton travel before reaching its turning point?

Your physics assignment is to figure out a way to use electricity to launch a small $6.0 - c m$ -long plastic drink stirrer. You decide that you’ll charge the little plastic rod by rubbing it with fur, then hold it near a long, charged wire, as shown in FIGURE P23.56. When you let go, the electric force of the wire on the plastic rod will shoot it away. Suppose you can uniformly charge the plastic stirrer to $10 n C$ and that the linear charge density of the long wire is $1.0 \times 10^{- 7} C / m$ . What is the net electric force on the plastic stirrer if the end closest to the wire is $2.0 c m$ away?

Hint: The stirrer cannot be modeled as a point charge; an integration is required.

Two 10 -cm-long thin glass rods uniformly charged to $+ 10 nC$ are placed side by side, $4.0 cm$ apart. What are the electric field strengths $E_{1}$ to $E_{3}$ at distances $1.0 cm, 2.0 cm$ , and $3.0 cm$ to the right of the rod on the left along the line connecting the midpoints of the two rods $?$ $?$

A $10 - cm$ long thin glass rod uniformly charged to $+ 10 nC$ and a $10 - cm$ -long thin plastic rod uniformly charged to $- 10 nC$ are placed side by side, $4.0 cm$ apart. What are the electric field strengths $E_{1}$ to $E_{3}$ at distances $1.0 cm, 2.0 cm$ , and from the glass rod a $3.0 cm$ long the line connecting the midpoints of the two rods?

See all solutions

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