Question

5. Charges of +20 u C each are kept at the corners A and B of a square ABCD each of side 9 cm and
charges of -20 u C each at the corners C and D. What is the total electrical potential energy of the
charges? Take infinity as the reference for zero PE. Ans ...... (-56.4 V)​ help me please......

Answer 1

Given :

• Charges +20µC are placed at A and B corners of a square ABCD.
• Charges -20µC are placed at corners C and D
• Side of square ABCD $\sf\:9cm=10{}^{-2}m$

To Find :

The total electrical potential energy of the charges .

Theory :

1)Electric potential energy due to two point charges :

${\purple{\boxed{\large{\bold{U=\dfrac{k\:q_{1}\times\:q_{2}}{r}}}}}}$

2)Electric potential energy due to system of point charges

${\red{\boxed{\large{\bold{U=U_{12} +U_{13} + U_{13}+ U_{14}.......}}}}}$

Solution :

Let the side of the square be r cm.

Then , AB=BC=CD=AD=r

And AC=BD =r√2

We have to find ,the total electrical potential energy of the charges.

$\sf\:U_{net}$

$=U_{12}+U _{13}+U_{14}+U _{23}+U_{24}+U_{34}$

$\sf=\dfrac{k\:q^2}{r}-\dfrac{k\:q^2}{r\sqrt{2}}-\dfrac{k\:q^2}{r}-\dfrac{k\:q^2}{r}-\dfrac{k\:q^2}{r\sqrt{2}}+\dfrac{k\:q^2}{r}$

$\sf=\dfrac{2k\:q^2}{r}-\dfrac{2k\:q^2}{r\sqrt{2}}-\dfrac{2k\:q^2}{r}$

$\sf=-\dfrac{2k\:q^2}{r\sqrt{2}}$

$\sf=-\dfrac{\sqrt{2}\:k\:q^2}{r}$

Now put the given values q=20µC and r = 9 cm

$\sf=\dfrac{-\sqrt{2}\times9\times\:10{}^{9}\times(20\times\:10{}^{-6})^2}{9\times10{}^{-2}}$

$\sf=\dfrac{-\sqrt{2}\times9\times10^9\times400\times10{}^{-12}\times10^2}{9}$

$\sf=\dfrac{-\sqrt{2}\times400}{10}$

$\sf=-40\sqrt{2}$

$\sf=-56.5V$

Therefore, the total electrical potential energy of the charges is -56.5V.

Answer 2

Answer:

Given :

Charges +20µC are placed at A and B corners of a square ABCD.

Charges -20µC are placed at corners C and D

Side of square ABCD \sf\:9cm=10{}^{-2}m9cm=10

−2

m

To Find :

The total electrical potential energy of the charges .

Theory :

1)Electric potential energy due to two point charges :

{\purple{\boxed{\large{\bold{U=\dfrac{k\:q_{1}\times\:q_{2}}{r}}}}}}

U=

r

kq

1

×q

2

2)Electric potential energy due to system of point charges

{\red{\boxed{\large{\bold{U=U_{12} +U_{13} + U_{13}+ U_{14}.......}}}}}

U=U

12

+U

13

+U

13

+U

14

.......

Solution :

Let the side of the square be r cm.

Then , AB=BC=CD=AD=r

And AC=BD =r√2

We have to find ,the total electrical potential energy of the charges.

\sf\:U_{net}U

net

=U_{12}+U _{13}+U_{14}+U _{23}+U_{24}+U_{34}=U

12

+U

13

+U

14

+U

23

+U

24

+U

34

\sf=\dfrac{k\:q^2}{r}-\dfrac{k\:q^2}{r\sqrt{2}}-\dfrac{k\:q^2}{r}-\dfrac{k\:q^2}{r}-\dfrac{k\:q^2}{r\sqrt{2}}+\dfrac{k\:q^2}{r}=

r

kq

2

−

r

2

kq

2

−

r

kq

2

−

r

kq

2

−

r

2

kq

2

+

r

kq

2

\sf=\dfrac{2k\:q^2}{r}-\dfrac{2k\:q^2}{r\sqrt{2}}-\dfrac{2k\:q^2}{r}=

r

2kq

2

−

r

2

2kq

2

−

r

2kq

2

\sf=-\dfrac{2k\:q^2}{r\sqrt{2}}=−

r

2

2kq

2

\sf=-\dfrac{\sqrt{2}\:k\:q^2}{r}=−

r

2

kq

2

Now put the given values q=20µC and r = 9 cm

\sf=\dfrac{-\sqrt{2}\times9\times\:10{}^{9}\times(20\times\:10{}^{-6})^2}{9\times10{}^{-2}}=

9×10

−2

−

2

×9×10

9

×(20×10

−6

)

2

\sf=\dfrac{-\sqrt{2}\times9\times10^9\times400\times10{}^{-12}\times10^2}{9}=

9

−

2

×9×10

9

×400×10

−12

×10

2

\sf=\dfrac{-\sqrt{2}\times400}{10}=

10

−

2

×400

\sf=-40\sqrt{2}=−40

2

\sf=-56.5V=−56.5V

Therefore, the total electrical potential energy of the charges is -56.5V.