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Chapter 19: Q14PE (page 696)

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50脳104V?

Short Answer

Expert verified

1.50脳106V/m is the required electric field strength.

Step by step solution

01

Principle

The magnitude of the electric field (E) at any point is given by the potential gradient at that point.

E=|-dVdr|

Here dV is the potential gradient and dr is the displacement from source charge.

02

The given data

  • The distance between the plates is:

d=(1.00cm)1m100cm=0.0100m

  • The potential difference between the two plates is:螖痴=1.50脳104V.
03

Calculation of electric field strength

Equation (1) is used to calculate the electric field strength E between the two plates:

E=螖痴d

After entering the numbers for 鈭�Vand d,

E=1.50脳104V0.0100m=1.50脳106V/m

Therefore, the electric field strength is1.50脳106V/m.

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Most popular questions from this chapter

How does the polar character of water molecules help to explain water's relatively large dielectric constant? (Figure 19.19)

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