1) What is Ex(P), the x-component of the electric field at point P, located at (
ID: 1580609 • Letter: 1
Question
1) What is Ex(P), the x-component of the electric field at point P, located at (x,y) = (8.5 cm, 0)? 2) What is a, the charge density on the surface of the conducting slab at x = 17 cm? 3) What is V(R) - V(P), the potentital difference between point P and point R, located at (x,y) = (8.5 cm, -21 cm)? 4) What is V(S) - V(P), the potential difference between point P and point S, located at (x,y) = (29.5 cm, -21 cm)? 4) What is V(S) - V(P), the potential difference between point P and point S, located at (x,y) = (29.5 cm, -21 cm)?5) What is Ex(T), the x-component of the electric field at point T, located at (x,y) = (46.5 cm, -21 cm)?
5) What is Ex(T), the x-component of the electric field at point T, located at (x,y) = (46.5 cm, -21 cm)? ViewProblem unititemiD 3332317&enrollmentio-; 309156 CSecure https/ GSpi, . Google Ace.. HSO Go nrenA aUL . Google Docs I of Infinite An infinite sheet of charge is located in the y-z plane at x-0 and has uniform charge denisity a 0.27 C/m2. Another infinite sheet of charge with unilform charge density O2- .39 pC/ ts located at x . c " 38 en. An unchanged nfinite conducting slab is placed haltway in between these sheets (i.e., between x 17 cm andx 21 cm). Insulato Insulator and Cond 1) What is E-P), the x-component of the electric field at point P, located at (x·y)·.5 cm, op? N/C Submi What is , the charge density on the surface of the conducting slab at x- 17 cm 3) what is veR-VIP), the potenttal difference between point P and point R, located at 0,y)-5 cm,-21 cm)? V Submit 4) what is visi . vn, the potential deference between point p and point S, located at C,y)·(29.5 crn,-21 cmi? MacBook Air ,3 3 4 5 6 9
Explanation / Answer
given
infinite charged sheet
in y-z plane at x = 0
sigma1 = 0.27 uC /m^2
sigma2 = -0.39 uC/m^2 at x = c = 38 cm
conducting slab between x = 17 to x = 21 cm
a. Ex at P (8.5,0) cm is given by
Ex = (sigma1 + sigma2)/epsilon = -6770.76 V/m
b. at x = 17 cm
charge on conducting slab should be sigma
then from gauss law
sigma = (sigma 1 + sigma2)*(-1)/2 = 0.06 uC/m^2
c. potential differnce between point P and R is 0 V
as electric field does not vary along y axis
d. V(s) - V(p) = V(s) - V(r) = 0 V ( as electric field is constant)
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