Correct, computer gets: 6.52E+06 N/C Hint: The electric field at position P due
ID: 1435252 • Letter: C
Question
Correct, computer gets: 6.52E+06 N/C
Hint: The electric field at position P due to charge Q1 is not influenced by charge Q2. Therefore, ignore charge Q2 and apply Coulomb's Law. Remember to convert all quantities to the appropriate SI unit.
b)What is the x-component of the total electric field at P?
Answer:
Hint: By the principle of linear superposition, the total electric field at position P is the vector sum of the electric field contribution from charge Q1 and Q2.
c)What is the y-component of the total electric field at P?
Answer:
d)What is the magnitude of the total electric field at P?
addition and recall that Ep = sqrt( Epx^2 + Epy^2 ).
Answer:
e)What is the angle (with respect to the x-axis) of the total electric field /
at point P (use deg as the symbol for degrees)?
Answer:
Explanation / Answer
A) x - Field = k(Q1+Q2)0.05/(0.05^2+0.035^2)^3/2 = 18.6*10^6 N/C
y - field = k(Q1 - Q2)0.035/(0.05^2+0.035^2)^3/2 = 5.54*10^6 N/c
Net field = 19.40 N/C
B) x - Field = k(Q1+Q2)0.05/(0.05^2+0.035^2)^3/2 = 18.6*10^6 N/C
C) y - field = k(Q1 - Q2)0.035/(0.05^2+0.035^2)^3/2 = 5.54*10^6 N/c
D) Net field = 19.40 N/C
E) theta = arctan(5.54/18.6) = 16.58 degrees