A charge of +8.0 mC is moved through a uniform electric field of 20 V/m. The electric field lines are horizontal and point to the left. Calculate the work done along path JKLM, where J to K is 2m to t

April 10, 2026Physics

ScanSolve AI Answer

April 10, 2026

Answer

20 V m^{-1}$.

Step 1: Identify the given values. The uniform electric field strength is $E = 20 V m^{-1}$ . The charge being moved is $q = +8.0 mC = +8.0 \times 10^{-3} \text{ C}$ . The electric field lines are horizontal and point to the left.

Step 2: Determine the work done along each segment of the path JKLM. The work done by a uniform electric field on a charge $q$ is given by $W = q \vec{E} \cdot \vec{d}$ , where $\vec{d}$ is the displacement vector. This can also be written as $W = q E d \cos\theta$ ,

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Identify the given values. The uniform electric field strength is E = 20 V m^-1.

Asked on April 10, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on April 10, 2026

Step 1: Identify the given values. The uniform electric field strength is E = 20 V m^-1. The charge being moved is q = +8.0 mC = +8.0 × 10^-3 C. The electric field lines are horizontal and point to the left. Step 2: Determine the work done along each segment of the path JKLM. The work done by a uniform electric field on a charge q is given by W = q E · d, where d is the displacement vector. This can also be written as W = q E d ,