Question

Electric field in space is given by \[ \vec{E} = 2x\hat{i} + 3y^2\hat{j} + 4\hat{k} \] A charge \(q = 3\,C\) is taken from \(r_1(0,-1,-3)\) to \(r_2(5,1,2)\). Find magnitude of \( \Delta U \).

Answer 1

Correct Answer: 47

Concept: Change in potential energy is related to work done by electrostatic field. \[ \Delta U = -W_{electric} \] Work done by electric field \[ W = q \int \vec{E}\cdot d\vec{r} \] Step 1: Compute line integral.} \[ \vec{E}\cdot d\vec{r} = 2x\,dx + 3y^2\,dy + 4\,dz \] \[ W = \int 2x\,dx + \int 3y^2\,dy + \int 4\,dz \]
Step 2: Integrate each term.} \[ \int 2x\,dx = x^2 \] \[ \int 3y^2\,dy = y^3 \] \[ \int 4\,dz = 4z \] Thus \[ W = (x^2 + y^3 + 4z) \]
Step 3: Evaluate between limits.} \[ W = (x^2 + y^3 + 4z)_{(0,-1,-3)}^{(5,1,2)} \] At \( (5,1,2) \) \[ 25 + 1 + 8 = 34 \] At \( (0,-1,-3) \) \[ 0 -1 -12 = -13 \] \[ W = 34 - (-13) \] \[ W = 47 \] Since \[ \Delta U = -W \] Magnitude \[ |\Delta U| = 47\,J \] Final Result \[ |\Delta U| = 47\,J \]