m-vb-va-kqab-dxx2-for-a-point-charge-on-the-x-axis-what-is-the-simplified-result

Question

For a point charge on the xx-axis, simplify the potential difference integral V(b)−V(a)=−kQ∫abdxx2\displaystyle V(b)-V(a)=-kQ\int_{a}^{b}\frac{dx}{$$x^2$$}.

Accepted Answer

kQ(1b−1a)\displaystyle kQ\left(\frac{1}{b}-\frac{1}{a}\right)

Answer

kQ(b−a)kQ(b - a)

Answer

kQ(1a−1b)\displaystyle kQ\left(\frac{1}{a}-\frac{1}{b}\right)

Answer

kQln⁡(ba)\displaystyle kQ\ln\left(\frac{b}{a}\right)

For a point charge on the xx-axis, simplify the potential difference integral V(b)−V(a)=−kQ∫abdxx2\(\displaystyle\) V(b)-V(a)=-kQ\(\int\)_{a}^{b}\(\frac{dx}{x^2}\).

For a point charge on the $x$ -axis, simplify the potential difference integral $V (b) - V (a) = - k Q \int_{a}^{b} \frac{d x}{x^{2}}$ .