consider a charge +q = 2.25 nc placed at (x = 0, y = 5.00 cm, z = 0). a line charge with linear charge…

consider a charge +q = 2.25 nc placed at (x = 0, y = 5.00 cm, z = 0). a line charge with linear charge density ρl = 2.50 nc/m is placed at x = 0 to x = 10.0 cm. determine the differential voltage at the position of +q due to the line charge, assuming that v(r → ∞) = 0 v.\n\na. dv = -\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}+y_{1}^{2}}}\n\nb. dv = +\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}-y_{1}^{2}}}\n\nc. dv = +\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}+y_{1}^{2}}}\n\nd. dv = -\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}-y_{1}^{2}}}\n\ndetermine the absolute potential at the position of +q.\nanswer:\n\ncalculate the work done in nj when moving +q from its initial position to (x = 10.0 cm, y = 5.00 cm, z = 0).\nanswer:\n\ncalculate the work done in nj when moving +q from its initial position to (x = 5.00 cm, y = 5.00 cm, z = 0).\nanswer:

consider a charge +q = 2.25 nc placed at (x = 0, y = 5.00 cm, z = 0). a line charge with linear charge density ρl = 2.50 nc/m is placed at x = 0 to x = 10.0 cm. determine the differential voltage at the position of +q due to the line charge, assuming that v(r → ∞) = 0 v.\n\na. dv = -\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}+y_{1}^{2}}}\n\nb. dv = +\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}-y_{1}^{2}}}\n\nc. dv = +\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}+y_{1}^{2}}}\n\nd. dv = -\\frac{1}{4\\pi\\epsilon_0}\\frac{dq}{\\sqrt{x^{2}-y_{1}^{2}}}\n\ndetermine the absolute potential at the position of +q.\nanswer:\n\ncalculate the work done in nj when moving +q from its initial position to (x = 10.0 cm, y = 5.00 cm, z = 0).\nanswer:\n\ncalculate the work done in nj when moving +q from its initial position to (x = 5.00 cm, y = 5.00 cm, z = 0).\nanswer:

Answer

Explanation:

Step1: Recall the formula for differential voltage

The differential voltage $dV$ due to a differential charge $dq$ at a distance $r$ is given by $dV=\frac{1}{4\pi\epsilon_0}\frac{dq}{r}$, where $\epsilon_0 = 8.85\times 10^{- 12}\ C^2/(N\cdot m^2)$. For a line - charge with linear charge density $\rho_L$, $dq=\rho_Ldx$. The distance between a point on the line - charge $(x,0,0)$ and the point $(0,y,0)$ is $r = \sqrt{x^{2}+y^{2}}$. So, $dV=\frac{1}{4\pi\epsilon_0}\frac{\rho_Ldx}{\sqrt{x^{2}+y^{2}}}$. The correct formula for the differential voltage at the position of $+Q$ due to the line - charge is $dV = +\frac{1}{4\pi\epsilon_0}\frac{dq}{\sqrt{x^{2}+y_1^{2}}}$ (option c).

Step2: Calculate the absolute potential

The absolute potential $V$ at the position of $+Q$ is obtained by integrating $dV$ over the length of the line - charge. [ \begin{align*} V&=\frac{\rho_L}{4\pi\epsilon_0}\int_{0}^{L}\frac{dx}{\sqrt{x^{2}+y^{2}}}\ &=\frac{\rho_L}{4\pi\epsilon_0}\left[\ln\left(x+\sqrt{x^{2}+y^{2}}\right)\right]_0^{L} \end{align*} ] where $L = 10.0\ cm=0.1\ m$ and $y = 5.00\ cm = 0.05\ m$, $\rho_L=2.50\ nC/m = 2.50\times 10^{-9}\ C/m$, $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$. [ \begin{align*} V&=9\times 10^{9}\times2.50\times 10^{-9}\left[\ln\left(0.1+\sqrt{0.1^{2}+0.05^{2}}\right)-\ln\left(0+\sqrt{0^{2}+0.05^{2}}\right)\right]\ &=22.5\left[\ln\left(0.1 + \sqrt{0.01 + 0.0025}\right)-\ln(0.05)\right]\ &=22.5\left[\ln\left(0.1+\sqrt{0.0125}\right)-\ln(0.05)\right]\ &=22.5\left[\ln\left(0.1 + 0.1118\right)-\ln(0.05)\right]\ &=22.5\left[\ln(0.2118)-\ln(0.05)\right]\ &=22.5\ln\left(\frac{0.2118}{0.05}\right)\ &=22.5\ln(4.236)\ &\approx22.5\times1.445\ &=32.5125\ V \end{align*} ]

Step3: Calculate the work done

The work done $W$ in moving a charge $Q$ through a potential difference $\Delta V$ is $W = Q\Delta V$.

Case 1: Moving from $(0,5,0)$ to $(10,5,0)$

The potential at $(0,5,0)$ is $V_1$ (calculated above $\approx32.5125\ V$). The distance from the line - charge at $(x = 10,y = 5)$ is the same as the distance from the line - charge at $(x = 0,y = 5)$ for this symmetric case. So, $\Delta V=0$ and $W_1=Q\Delta V = 2.25\times 10^{-9}\times0=0\ nJ$.

Case 2: Moving from $(0,5,0)$ to $(5,5,0)$

We need to calculate the potential at $(x = 5,y = 5)$. First, we calculate the potential $V_2$ at this point using the integral formula $V=\frac{\rho_L}{4\pi\epsilon_0}\int_{0}^{L}\frac{dx}{\sqrt{(x - 5)^{2}+y^{2}}}$ (using a change of variables to account for the new position). But we can also use the super - position principle. The potential difference $\Delta V$ between the two points: The potential at $(0,5)$ is $V_1$ and at $(5,5)$: [ \begin{align*} V_2&=\frac{\rho_L}{4\pi\epsilon_0}\int_{0}^{0.1}\frac{dx}{\sqrt{(x - 0.05)^{2}+0.05^{2}}}\ \end{align*} ] Let $u=x - 0.05$, $du=dx$. When $x = 0$, $u=-0.05$ and when $x = 0.1$, $u = 0.05$. [ \begin{align*} V_2&=\frac{\rho_L}{4\pi\epsilon_0}\int_{-0.05}^{0.05}\frac{du}{\sqrt{u^{2}+0.05^{2}}}\ &=\frac{\rho_L}{4\pi\epsilon_0}\left[\ln\left(u+\sqrt{u^{2}+0.05^{2}}\right)\right]_{-0.05}^{0.05}\ &=\frac{\rho_L}{4\pi\epsilon_0}\left[\ln\left(0.05+\sqrt{0.05^{2}+0.05^{2}}\right)-\ln\left(-0.05+\sqrt{0.05^{2}+0.05^{2}}\right)\right]\ &=9\times 10^{9}\times2.50\times 10^{-9}\left[\ln\left(0.05+\sqrt{0.005}\right)-\ln\left(-0.05+\sqrt{0.005}\right)\right]\ &=22.5\left[\ln\left(0.05 + 0.0707\right)-\ln\left(-0.05+0.0707\right)\right]\ &=22.5\left[\ln(0.1207)-\ln(0.0207)\right]\ &=22.5\ln\left(\frac{0.1207}{0.0207}\right)\ &=22.5\ln(5.83)\ &\approx22.5\times1.763\ &=39.6675\ V \end{align*} ] $\Delta V=V_2 - V_1=39.6675 - 32.5125 = 7.155\ V$ $W_2=Q\Delta V=2.25\times 10^{-9}\times7.155\times10^{9}=16.09875\ nJ\approx16.10\ nJ$

Answer:

  1. Option: C. $dV = +\frac{1}{4\pi\epsilon_0}\frac{dq}{\sqrt{x^{2}+y_1^{2}}}$
  2. $32.51\ V$
  3. $0\ nJ$
  4. $16.10\ nJ$