If the electric field intensity between two charged parallel plates have the magnitude 2 x 10^2 N/C, then the magnitude of electrostatic force on an electron in the electric field willbe(Note: The charge on the electron is equal to 1.6 x 10^-19 C.)
Correct answer: B. 3.2 x 10^-17 N
- A. 8.0 x 10^-22 N
- B. 3.2 x 10^-17 N
- C. 2.0 x 10^2 N
- D. 1.2 x 10^21 N
Explanation
The correct answer is Option B: 3.2 × 10-17 N. To calculate the electrostatic force on an electron, we use the formula F = qE, where q is the charge of the electron (1.6 × 10^(-19) C) and E is the electric field intensity (2 × 10^2 N/C). Substituting these values, we get F = (1.6 × 10^(-19) C) * (2 × 10^2 N/C) = 3.2 × 10^(-17) N. Therefore, the magnitude of the electrostatic force on an electron in the electric field is 3.2 × 10^(-17) N. The other options either incorrectly calculate the force or provide unrelated values.
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About Electric Field Intensity
Electric field intensity is the force acting per unit positive test charge, with direction set by the force on that charge. Work includes fields of point charges, vector superposition, field lines and the relation between field and electric potential. Field intensity must be distinguished from electric force and potential.
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