A charge of \(4.6 \times 10^{-5}\) C is placed in an electric field of intensity \(3.2 \times 10^{4}\ \mathrm{V\,m^{-1}}\). What is the force acting on the ...
Question 1 Report
A charge of \(4.6 \times 10^{-5}\) C is placed in an electric field of intensity \(3.2 \times 10^{4}\ \mathrm{V\,m^{-1}}\). What is the force acting on the electron?
Answer Details
To calculate the force acting on the charge in an electric field, we can use the formula:
F = q * E
Where:
F is the force acting on the charge,
q is the charge of the particle, and
E is the electric field intensity.
In this case, the charge is given as 4.6 × 10^(-5) C and the electric field intensity is given as 3.2 × 10^4 V/m.
Substituting these values into the formula:
F = (4.6 × 10^(-5) C) * (3.2 × 10^4 V/m)
To multiply numbers in scientific notation, we multiply the coefficients and add the exponents:
F = (4.6 * 3.2) * (10^(-5 + 4)) C * V/m
F = 14.72 * 10^(-1) C * V/m
To simplify, we can convert the result to standard form:
F = 1.472 C * V/m
Therefore, the force acting on the charge is **1.472 N**.
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