A particle carrying (+3.00 mathrm{nC}) of positive charge is at the origin of an (x y) coordinate

Question:

A particle carrying \(+3.00 \mathrm{nC}\) of positive charge is at the origin of an \(x y\) coordinate system. Take \(V(\infty)\) to be zero for these calculations.

(a) Calculate the potential \(V\) on the \(x\) axis at \(x=3.0000 \mathrm{~m}\) and at \(x=3.0100 \mathrm{~m}\).

(b) Does the potential increase or decrease as \(x\) increases? Compute \(\Delta V / \Delta x\) for distances along the \(x\) axis, where \(\Delta V\) is the difference in potential between \(x=3.0000 \mathrm{~m}\) and \(x=3.0100 \mathrm{~m}\), and \(\Delta x=0.0100 \mathrm{~m}\).

(c) Calculate the electric field magnitude at \(x=3.0000 \mathrm{~m}\), and compare this magnitude with the value calculated for \(\Delta V / \Delta x\) in part \(b\).

(d) Calculate the electrostatic potential at the position \(x=3.0000 \mathrm{~m}, y=0.0100 \mathrm{~m}\), and compare your result with the potential at \(x=3.0000 \mathrm{~m}\) calculated in part \(a\). Discuss the significance of this result.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: