Find the charge\'s location. The electric field at the point z = 5.00 cm and y 0

Question

Find the charge's location. The electric field at the point z = 5.00 cm and y 0 points in the positive r direction with a magnitude of 8.00 N/C. At the point z = 10.0 cm and y = 0 the electric field points in the positive z direction with a magnitude of 16.0N/C.Assume this electric field is produced by a single point charge cm Submit XIncorrect: Try Again; 9 attempts remaining Part B Find the sign of the charge positive negative Previous Ans Correct Part C Find the magnitude of the charge pC Submit

Explanation / Answer

We see from the given condition that we have two different values for Electri field (E) at two different positions which means we have got two equations with two unknowns, which are x, the position on the line of charge Q, and the magnitude of Q.

we know that the electric field E created by a charge Q at distance d from Q is given by the equation

E = (k* Q) / d2  --------------------(1)

where:
k = 9*109 N*m2/ C2

Q is the point charge creating the field E

d = the distance in meters between Q and a test charge of 1C




If E1 is the value (=8) at the point x = 0.0500m and
E2 is the value (=16) at the point x = 0.100m,
then we have the two equations:

E1 = k * Q / (x - 0.0500)2 = 8 N/C
and
E2 = k * Q / (x - 0.100)2 = 16 N/C

Multiplying both sides by the denominator:

k * Q = 8 * (x - 0.0500)2

k * Q = 16 * (x - 0.100)2

Since both the right sides equal the same left side, we can equate the two left sides:

8 * (x - 0.0500)2 = 16 * (x - 0.100)2  

or 8 * (x2 - 0.100x + 0.0025) = 16 * (x2 - 0.200x + 0.0100)  

0r 8 x2 - 0.8x + 0.02 = 16 x2 - 3.2x + 0.16

Subtracting the left side from both sides:

0 = 8 x2 - 2.4x + 0.14   

this is a quadratic equation, we solve this equation to get the value of x

solving, we get x = 22.1 cm, 7.93 cm

we can rule out x = 7.93 cm, because, if Q were placed there, the electric field E2 direction would be the reverse of E1, which is not true

therefore we have x = 0.221 m as the location for Q

now

E1 = k* Q / (x - 0.0500)2 = 8

Solving for Q we get:

Q = (x - 0.0500)2 * 8 / k

= (0.221 - 0.0500)2 * 8 / (9*109)

= 0.026*10-9 C = 26*10-12 C = 26 pC

Since the direction of E1 is toward +x, Q will have to be negative (so that it will attract a positive charge placed at 5.00cm:

therefore Q = - 26 pC

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.