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Difficulty: EasyElectrostatics and Electric Charges

An electric charge of +3.2×10−19 C+3.2 \times 10^{-19}\text{ C} is situated in a uniform electric field of strength 5.0×104 N C−15.0 \times 10^4\text{ N C}^{-1}. What is the magnitude of the electrostatic force exerted on the charge?

  1. A
    6.4×10−24 N6.4 \times 10^{-24}\text{ N}
  2. 1.6×10−14 N1.6 \times 10^{-14}\text{ N}Answer
  3. C
    1.6×10−23 N1.6 \times 10^{-23}\text{ N}
  4. D
    5.0×104 N5.0 \times 10^4\text{ N}

Answer

The magnitude of the electrostatic force exerted on the charge is 1.6×10−14 N1.6 \times 10^{-14}\text{ N}.
The relationship between electric field intensity EE, charge qq, and electrostatic force FF is given by F=qEF = qE. Multiplying +3.2×10−19 C+3.2 \times 10^{-19}\text{ C} by 5.0×104 N C−15.0 \times 10^4\text{ N C}^{-1} gives 1.6×10−14 N1.6 \times 10^{-14}\text{ N}, which correctly expresses the force magnitude in standard notation.

Step-by-Step Solution

1
Identify the given values and formula.
Charge q=3.2×10−19 Cq = 3.2 \times 10^{-19}\text{ C}, Electric field strength E=5.0×104 N C−1E = 5.0 \times 10^4\text{ N C}^{-1}. Formula: F=qEF = qE.
The force experienced by a charge in an electric field is the product of the charge magnitude and the field strength.
2
Substitute the values into the formula and calculate.
F=(3.2×10−19)×(5.0×104)=16.0×10−15 N=1.6×10−14 NF = (3.2 \times 10^{-19}) \times (5.0 \times 10^4) = 16.0 \times 10^{-15}\text{ N} = 1.6 \times 10^{-14}\text{ N}.
Multiplying the numerical coefficients (3.2×5.0=16.0)(3.2 \times 5.0 = 16.0) and combining the powers of ten (10−19×104=10−15)(10^{-19} \times 10^4 = 10^{-15}) gives 1.6×10−14 N1.6 \times 10^{-14}\text{ N} in standard scientific notation.

Key Concept

Electric Field Intensity and Electrostatic Force (F=qEF = qE)
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Topics

Electrostatics and Electric Charges

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