AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the diagram,
By using triangle property, the distance AC is,
AC=√(AB)2+(BC)2

AC=5cm

AC=5×10−2m

The initial electric potential energyof the three charges is,

U=‌

kq1q2

AB

+‌

kq1q3

AC

+‌

kq2q3

BC

U=k[‌

q1q2

AB

+‌

q1q3

AC

+‌

q2q3

BC

]

Substitute values.
U=(9×109)[‌

4×3×10−12

4×10−2

+‌

4×5×10−12

5×10−2

+‌

3×5×10−12

3×10−2

]

U≈10.8J
When the three charges located on an equilateral triangle of side 3cm as shown in figure below,
The final potential energy is calculated as,
U′=(9×109)[‌

4×3×10−12

3×10−2

+‌

4×5×10−12

3×10−2

+‌

3×5×10−12

3×10−2

]
U′≈14.1J
The work done in moving charges at points A and C is
calculated as,
W=U′−U
W=14.1−10.8
W=3.3J