Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
The values of the potential differences for the three questions are;
A) [tex]V_{MN} = -147 V[/tex]
B) [tex]V_{MQ}[/tex] = -120 V
C) [tex]V_{N} = 51 V[/tex]
We are given the expression of the electric field as;
E = (6x² x^ + 6y y^ +4 z^) V/m
[tex]V_{MN} = -\int\limits^M_N {E} \, dx[/tex]
Integrating this with the M and N coordinates as boundaries in mind gives;
[tex]V_{MN} = -[6\frac{x^{3}}{3} + 6\frac{y^{2}}{2} + 4z]^{2,6,1}_{-3,-3,2}[/tex]
[tex]V_{MN} = -[2{x^{3} + 3y^{2} + 4z]^{2,6,1}_{-3,-3,2}[/tex]
Plugging in those boundary values and solving using an online integral calculator gives;
[tex]V_{MN} = -147 V[/tex]
[tex]V_{MQ} = -[6\frac{x^{3}}{3} + 6\frac{y^{2}}{2} + 4z]^{2,6,1}_{4,-2,-35}[/tex]
[tex]V_{MQ} = -[2{x^{3} + 3y^{2} + 4z]^{2,6,1}_{4,-2,-35}[/tex]
Plugging in those boundary values and solving using an online integral calculator gives;
[tex]V_{MQ}[/tex] = -120 V
C) We are told that V = 2 at P(1,2,4). Thus potential difference between point V and N is;
[tex]V_{NP} = -[6\frac{x^{3}}{3} + 6\frac{y^{2}}{2} + 4z]^{-3,-3,2}_{1,2,4}[/tex]
[tex]V_{NP} = -[2{x^{3} + 3y^{2} + 4z]^{-3,-3,2}_{1,2,4}[/tex]
Plugging in those boundary values and solving using an online integral calculator gives; [tex]V_{NP} = 49 V[/tex]
Thus;
[tex]V_{N} = V + V_{NP}[/tex]
[tex]V_{N}[/tex] = 2 + 49
[tex]V_{N} = 51 V[/tex]
Read more about Electric field vectors at; https://brainly.com/question/13193669
