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

AI Thread Summary
The electric field measurements indicate a point charge is responsible for the varying magnitudes at two locations. The equations derived from the electric field values suggest that the charge is negative and located beyond x = 20 cm. However, calculations for the charge's location and magnitude have yielded incorrect results despite attempts to solve them. A critical note is made to measure distances in meters rather than centimeters for accurate calculations. Assistance is requested to resolve the discrepancies in the findings.
ajmCane22
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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.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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