Where is the point charge located and what is its charge in this electric field?

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SUMMARY

The electric field measurements at points (10.0 cm, 0) and (20.0 cm, 0) indicate a point charge is responsible for the observed electric field strengths of 10.0 N/C and 20.0 N/C, respectively. The calculations reveal that the charge is located at approximately 24.14 cm along the x-axis and has a negative charge of -0.13 nC. The error in previous attempts stemmed from using centimeters instead of meters for distance measurements, which is crucial for accurate calculations in electrostatics.

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Homework Statement



The electric field at the point x = 10.0 cm and y = 0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x = 20.0 cm and y = 0 the electric field points in the positive x direction with a magnitude of 20.0 N/C. Assume that this electric field is produced by a single, point charge.

(a) Find the location of the point charge.

(b) Find the sign and magnitude of its charge.

The Attempt at a Solution



10 =q/4piε(d+10)^2

20 = q/4piε(d)^2
E is more at x=20 , so q must be -ve, and situated beyond x=20

dividing both eqn , we get
20/10 = d^2 + 100+ 20d/ d^2
solving the quadratic eqn , we get
d= 10±10√2
d= 24.14cm
put d= 24.14 in any above formula,

we get q= -0.13nC

Both answers are incorrect. I've gone to two different tutors and neither could help. Please help
 
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Measure distances in meters, not cm.
 

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