Two point charges, q1 and q2 are separated by a distance d. For convenience, we will say that q1 is at the origin (0,0) on a coordinate system, q2 is at (d,0) and a reference point p is directly above q2 at a distance h at point (d,h). q1 has value 1μ C and q2 has value 8μ C. Take d= 1.1m and h= 0.7m. Note k=9×109Nm2/C2What is the strength of the electric field at point p in N? Give your answer to three significant digits and do not include the units.
Question
Two point charges, q1 and q2 are separated by a distance d. For convenience, we will say that q1 is at the origin (0,0) on a coordinate system, q2 is at (d,0) and a reference point p is directly above q2 at a distance h at point (d,h). q1 has value 1μ C and q2 has value 8μ C. Take d= 1.1m and h= 0.7m.
Note k=9×109Nm2/C2What is the strength of the electric field at point p in N? Give your answer to three significant digits and do not include the units.
Solution
To find the electric field at point P, we need to find the electric field due to each charge at point P and then add them vectorially.
- Electric field due to q1 at P (E1):
The distance from q1 to P is the hypotenuse of a right triangle with sides d and h. So, r1 = sqrt(d^2 + h^2) = sqrt((1.1m)^2 + (0.7m)^2) = 1.3m.
The electric field due to a point charge is given by E = kq/r^2. So, E1 = kq1/r1^2 = (9×10^9 Nm^2/C^2)*(1μC)/(1.3m)^2 = 5.3 N/C.
The direction of E1 is along the line from q1 to P, or in the positive y direction.
- Electric field due to q2 at P (E2):
The distance from q2 to P is h = 0.7m.
So, E2 = kq2/h^2 = (9×10^9 Nm^2/C^2)(8μC)/(0.7m)^2 = 145.7 N/C.
The direction of E2 is along the line from q2 to P, or in the negative y direction.
- The total electric field at P (E):
Since E1 and E2 are in opposite directions, we subtract their magnitudes to find the total electric field: E = E2 - E1 = 145.7 N/C - 5.3 N/C = 140.4 N/C.
So, the strength of the electric field at point P is 140 N/C to three significant digits.
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