Calculating Electric Field Strength: +5.0810^-6 Charge at 0.3m from Sphere

AI Thread Summary
To calculate the electric field strength 0.3m from a small metal sphere carrying a charge of +5.0810^-6, the relevant formula is E = kq/d^2, where k is Coulomb's constant. Initially, there was confusion regarding the use of two charges in the calculation, but it was clarified that only one charge is needed. The user successfully resolved the issue by applying the correct formula. The discussion emphasizes understanding the definition and application of electric field strength in relation to point charges. Overall, the key takeaway is the correct application of the electric field formula for a single charge.
pat666
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Homework Statement


calculate the electric field strength 0.3m from the centre of a small metal sphere cArying a charge of +5.0810^-6


Homework Equations





The Attempt at a Solution


TRIED putting into f = q1q2k/d^2 with q1 and q2 both being the known charge, but did not work

any help is appreciated thanks
 
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How you define electric field?
 
i don't know what you mean the entire question is written out.
 
pat666 said:
i don't know what you mean the entire question is written out.
What is the expression for an electric field at a point d from a charge q?
In the problem only one charge is given.
 
not sure, that's the question given. maybe its not right.
 
dont worry i figured it out myself it was E = kq/d^2
thanks anyway.
 
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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