Question

Pls Help Id rlly appreciate it thx!!

Answer 1

Given:-

`\rm \: E_x \: = 250 \: N/C`

`\rm \: q \: = 18 \: mC = 18 * {10}^( - 6) C(along \: x \: axis)`

`\rm \: d = 20 \: cm = 0.2 \: m \: (along \: x \: axis)`

`\rm \: d = 60\: cm = 0.6 \: m \: (along \: y\: axis)`

Unknown

`\rm Potential \: difference (\Delta V)`

Then

`\rm \: Potential \: energy =( \Delta U )`

Formula

`\rm \Delta V = - E_x.d_x`

`\rm \: \Delta U = q. \Delta V`

Solution

`\rm \Delta V = - E_x.d_x`

`\rm \Delta V = - 250 * 0.2`

`\rm \Delta V = - 250 * (2)/(10) = - 50V`

Again

`\rm \: \Delta U = q. \Delta V`

`\rm \: \Delta U = 18 * {10}^( - 6) * ( - 50)`

`\rm \: \Delta U = 18 * {10}^( - 6) * - ( 0.5 * {10}^(2) )`

`\rm \: \Delta U = - 9 * {10}^( - 4)J`

`\rm \: Potential \: energy = \Delta U = - 9 * {10}^( - 4)J`