Question

find the work done by an external agent in slowly shifting a charge q = 1μC in the electric field E =10^(3iˆ)V/m from the point P(1,2) to Q(3,4).

Answer 1

Given :

Magnitude of charge = 1μC

Electric field (E) = 10³i V/m

To Find :

Work done in shifting the charge

Solution :

• Work done against the electric field is    $\int\limits^2_1 {-q(Ei)} \, (dxi + dyj)$

•    W  =      $\int\limits^2_1 {-q(Ei)} \, (dxi + dyj)$

•    W =       $\int\limits^2_1{-qE} \, dx$

•    W =      -qE ( x₂ - x₁ )

•    W =      -[10⁻⁶× 10³ ×(3-1)]

•    W =      -2 × 10⁻³ J

Work done in shifting the charge is  -2 × 10⁻³ J

Answer 2

The work done by an external agent in slowly shifting a charge in the electric field is -2 × 10⁻³ J

Given:

q = 1 μC

E = $10^{3\vec{l}}$ V/m

Explanation:

The work done against the electric field is given by the formula:

W = -q $\int_{1}^{2} \vec{E} \vec{dl}$

W = -q ∫ (E $\vec{i}$)(dx $\vec{i}$ + dy $\vec{j}$)

W = -q $\int_{x_{1}}^{x_{2}}$ E.dx

W = -qE (x₂ - x₁)

On substituting the values, we get,

W = -10⁻⁶ × 10³ × (3 - 1)

W = -2 × 10⁻³ J