A charge of 4.6×10−5 4.6 × 10 − 5 C is placed in an electric field of intensity 3.2×104 3.2 × 10 4 Vm−1 . What is the force acting on the electron?

A charge of 4.6×10−5 $4.6\times {10}^{-}5$C is placed in an electric field of intensity 3.2×104 $3.2\times {10}^4$ Vm−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**.