Is the Electric Field Calculation Consistent with the Potential Result?

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Cactus
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Homework Statement
Hey, I just wanted to post these two questions to check whether the answers I've gotten are correct, as I have no real way of checking if what I've done is correct (This is for parts c and d of the attached question)

For c, would the electric field along the z axis just be the superposition of the field from a negatively charged plate plus the field of a positively charged sphere (as along the z axis the field lines are parallel to the axis)

Likewise, for d, would this be the correct way to go about solving for capacitance given the formula for potential, as I can't see any other way to cancel the Vo and express capacitance purely in geometric terms
Relevant Equations
Capacitance = Charge/Potential Difference
Electric Field of an Infinite Plate
Electric Field of a Sphere
Question
Diagram of a solid conducting sphere of radius R labelled +Q, centred a height d above a square conducting plate labelled -Q, with a z-axis pointing upward and d marked as the vertical distance between them.


Part C
Handwritten derivation on lined paper: a sphere of radius R with charge +Q and a plane sheet with charge -Q, point P at distance x from the plate, with r = R + d - x, E+ = Q/(4πε0 r^2), E- = |σ|/(2ε0) = Q/(2ε0 A), and total field E_net = Q/(2ε0)(1/(2π(R+d-x)^2) + 1/A).


Part D
Handwritten derivation on lined paper of the capacitance C = Q/ΔV for a parallel setup, computing ΔV from potential terms with 1/(2πR), 1/(2πdR) and d-R over A, reducing to C = 2ε₀ / ((d-R)(1/(2πdR) + 1/A)).
 
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at part C How you can write $$ E_{-}=\frac{|\pi|}{2\epsilon_0} $$ and what is $$ \pi $$
 
Masano Hutama said:
at part C How you can write $$ E_{-}=\frac{|\pi|}{2\epsilon_0} $$ and what is $$ \pi $$
ah that's not pi its just a badly written n as in surface charge density
 
Part c must be correct from you, only thing I have to say is that the d in the book scheme and the d in your notes are different. If I call d' the d of your notes and simply d the d from your book it is

d=d'+R, where R is the radius of the solid sphere,

and having that in mind, the result you get for electric field E in part c seems to be in agreement with the result for potential V that is given by the book(Just integrate your E-field and you ll get the V as presented by the book, we know that ##V=\int Edr##).

Part d also seems correct.
 
Delta2 said:
Part c must be correct from you, only thing I have to say is that the d in the book scheme and the d in your notes are different. If I call d' the d of your notes and simply d the d from your book it is

d=d'+R, where R is the radius of the solid sphere,

and having that in mind, the result you get for electric field E in part c seems to be in agreement with the result for potential V that is given by the book(Just integrate your E-field and you ll get the V as presented by the book, we know that ##V=\int Edr##).

Part d also seems correct.
Yeah I realized my mistake with the d and d' after posting this and fixed that, but thanks for the reply and confirmation on answers. I've also had a friend finish this question now and got the same answers so fingers crossed they're right