Question

A +4.0 uC charge is placed on the x axis at x= +3.0 m, and a -2.0 uC is located on the y-axis at y= -1.0 m. Point A is on the y axis at y= +4.0 m. Determine the electric potential at point A (relative to zero at the origin).

Answer 1

Answer:

The potential is  $V_A = 9600 \ V$

Explanation:

From the question we are told that  

   The  magnitude of the charge is  $q_1 = 4 \mu C = 4*10^{-6} \ C$

   The position of the charge is  $x = + 3.0 \ m$

   The magnitude of the second charge is  $q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C$

   The position is  $y_1 = - 1.0 \ m$

     The position of point A is  $y_2 = + 4.0 \ m$  

Generally the electric potential  at A due to the first charge is mathematically represented as

         $V_a = \frac{k * q_1 }{r_1 }$

Here k is the coulombs constant with value $k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}$

        $r_1$  is the distance between first charge and a which is mathematically represented as

         $r_1 = \sqrt{x^2 + y_2 ^2 }$

=>      $r_1 = \sqrt{3^2 + 4 ^2 }$    

=>      $r_1 = 5 \ m$    

So

           $V_a = \frac{9*10^9 * 4*10^{-6} }{5 }$

      $V_a = 7200 \ V$

Generally the electric potential  at A due to the second charge is mathematically represented as

         $V_b = \frac{k * q_2 }{r_2 }$

Here k is the coulombs constant with value $k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}$

        $r_2$  is the distance between second charge and a which is mathematically represented as

         $r_2 = y_2 - y$

=>      $r _2 = 4.0 - (-1.0)$    

=>      $r = 5 \ m$    

So

           $V_a = \frac{9*10^9 * -2*10^{-6} }{5 }$

      $V_a = -3600 \ V$

So the net potential difference at point A  due to the charges is mathematically represented as

       $V_n = V_a + V_b$

=>     $V_n = 7200 - 3600$

=>     $V_n = 3600 V$

Generally the net potential difference at the origin due to both charges is mathematically represented as

     $V_N = V_c + V_d$

Here  

      $V_c = \frac{k * q_1 }{x}$

=>   $V_c = \frac{9*10^9 * 4*10^{-6} }{3}$

=>   $V_c = 12000 V$

and

              $V_d= \frac{k * q_2 }{y}$

=>   $V_c = \frac{9*10^9 * -2*10^{-6} }{1}$

=>   $V_c =- 18000 V$

Generally the net potential difference at the origin is  

       $V_N = 12000 - 18000$

=>     $V_N = -6000$

Generally the potential difference at A relative to zero at the origin is mathematically evaluated as

         $V_A = V_n - V_N$

=>      $V_A = 3600 - (-6000)$

=>      $V_A = 9600 \ V$