Question

ANSWER ALL PLEASE
Question 16 2 pts In the the figure that is shown below, the y-intercept is (0,36) and the x-intercept is (6, 0). Find the electric potential difference between x= 0 andx=6 m. Note that they and z components of the electric field are zero. If your answer is negative, put the minus sign in the answer box. Please round your answer to one decimal place. E: (N/C) (0.y) (x,0) * (m) Equation: V=- | E (x) dx Integrate from x=0 to x = 6. Hint: Find the slope of the straight line (which is negative) and set E (x)-36 = slope (2)

Answer 1

13. Particles A, B and D, since,

A = 9.6*10^-19 C = 6e, B = 6.4*10^-19 C = 4e and

D = 5e

17. We have,

  $\Delta V=-\int E(x)dx$

Here, E_x = 4x, which is to integrate from 2 to 4

E_y = -20 y², which is to integrate from 3 to 6

$\Delta V=- 4\int_{2}^{4}xdx +20\int_{3}^{6}y^2dy$

= -4[4²/2 - 2²/2] + 20[6³/3 - 3³/3]

= -24 + 1260

= 1236 volt

16. Slope, m = y/x = -(36/6) = - 6 N/C.m

Y intercept, c = 36 N/C

Equation of the straight line, y = mx + c, so,

E(x) = -6x + 36

Now, we have,

$\Delta V=-\int{E(x)}{dx}$

$=-\int_{0}^{6}{(-6x+36)}{dx}$

$=6\int_{0}^{6}{x}{dx}-36\int_{0}^{6}{dx}$

= 6*[6²/2 - 0²/2] - 36*[6-0]

= 108 - 216

= - 108 volt