Hard question need help :)?

Question

Two point charges q1 = -1.03 μC and q2 = 3.63 μC are located at two corners of the rectangle with side lengths a = 24.1 cm and b = 47.5 cm, as shown in the figure. 
a) What is the electric potential at point A? 
b)What is the potential difference between points A and B? 
ty for help

Answer 1

(a)

Two point charges q1 = -1.03 μC and q2 = 3.63 μC .

The electric potential at point A due to point charge q1 is 

                      V₁ = (1/4πƐ₀)(q1/r )

Where q1 is  point charge . ( q1 = -1.03 μC )

            r is the distance between point charge and the point  where we need to find the electric potential .

                  ( a = r = 24.1 cm = 0.241 m )

           Ɛ₀ is constant value . (   Ɛ₀ = 8.854*10^-12  )

 V₁ = (_.1/4π(8.854*10^-12))( -1.03*10^-6 / 0.241 )

 V₁ = (0.898 * 10^10 )( -4.273 * 10^-6 )

 V₁ = -3.841 * 10^4  J/C .

The electric potential at point A due to point charge q2 is 

                      V₁ = (1/4πƐ₀)(q2/r )

Where q2 is  point charge . ( q2 =3.63 μC )

            r is the distance between point charge and the point  where we need to find the electric potential .

                                       ( b = r = 47.5 cm = 0.475 m )

           Ɛ₀ is constant value . (   Ɛ₀ = 8.854*10^-12  )

 V₂ = (_.1/4π(8.854*10^-12))( 3.63*10^-6 / 0.475 )

 V₂ = (0.898 * 10^10 )( 7.642 * 10^-6 )

 V₂ =6.868 * 10^4 J/C .

So the resultant electric potential at point A  V_A  =  V₁ + V₂

V_A  =   -3.841 * 10^4 +6.868 * 10^4 

V_A  = 3.027 * 10^4  J/C 

So the electric potential at point A  is  3.027 * 10^4  J/C  .

Answer 2

(b)

Two point charges q1 = -1.03 μC and q2 = 3.63 μC .

The electric potential at point A due to point charge q1 amd q2 is 

V_A  =  (1/4πƐ₀)(q1/r1 ) + (1/4πƐ₀)(q2/r2 )

Where  

q1 is  point charge . ( q1 = -1.03 μC )

q2 is  point charge . ( q2 =3.63 μC )

r1 is the distance between point charge q1 and the point A . ( r1 = 24.1 cm = 0.241 m )

r2 is the distance between point charge q2 and the point A . ( r2 = 47.5 cm = 0.475 m )

 Ɛ₀ is constant value . (   Ɛ₀ = 8.854*10^-12  ) 

V_A  =  (1/4πƐ₀)(q1/r1 ) + (1/4πƐ₀)(q2/r2 )

V_A  =  (1/4π(8.854*10^-12))( -1.03*10^-6 / 0.241 ) + (1/4π(8.854*10^-12))( 3.63*10^-6 / 0.475 )

V_A  =  (0.898 * 10^10 )( -4.273 * 10^-6 ) + (0.898 * 10^10 )( 7.642 * 10^-6 )

V_A  =   -3.841 * 10^4 +6.868 * 10^4 

V_A  = 3.027 * 10^4  J/C 

So the electric potential at point A  is  3.027 * 10^4  J/C .

The electric potential at point B due to point charge q1 amd q2 is 

V_A  =  (1/4πƐ₀)(q1/r1 ) + (1/4πƐ₀)(q2/r2 )

Where  

q1 is  point charge . ( q1 = -1.03 μC )

q2 is  point charge . ( q2 =3.63 μC )

r1 is the distance between point charge q1 and the point B . ( r1 = 47.5 cm = 0.475 m  )

r2 is the distance between point charge q2 and the point B . ( r2 = 24.1 cm = 0.241 m )

 Ɛ₀ is constant value . (   Ɛ₀ = 8.854*10^-12  ) 

V_B  =  (1/4πƐ₀)(q1/r1 ) + (1/4πƐ₀)(q2/r2 )

V_B =  (1/4π(8.854*10^-12))( -1.03*10^-6 / 0.475  ) + (1/4π(8.854*10^-12))( 3.63*10^-6 / 0.241 )

V_B  =  (0.898 * 10^10 )( -2.168 * 10^-6 ) + (0.898 * 10^10 )( 15.062 * 10^-6 )

V_B  =   -1.948 * 10^4 +13.537 * 10^4 

V_B  = 11.589 * 10^4  J/C 

The potential difference between points A and B due to B is V = V_B - V_A 

V = 11.589 * 10^4  J/C - 3.027 * 10^4  J/C

V = 8.562 * 10^4  J/C .

So the potential difference between points A and B due to B is   8.562 * 10^4  J/C .