23.23 a uniform electric field has magnitude e and is directed in the negative x - direction. the potential…

23.23 a uniform electric field has magnitude e and is directed in the negative x - direction. the potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) is 240 v. (a) which point, a or b, is at the higher potential? (b) calculate the value of e. (c) a negative point charge q = - 0.200 μc is moved from b to a. calculate the work done on the point charge by the electric field.

23.23 a uniform electric field has magnitude e and is directed in the negative x - direction. the potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) is 240 v. (a) which point, a or b, is at the higher potential? (b) calculate the value of e. (c) a negative point charge q = - 0.200 μc is moved from b to a. calculate the work done on the point charge by the electric field.

Answer

Explanation:

Step1: Determine higher - potential point

The electric field $\vec{E}$ points in the direction of decreasing electric potential. Since $\vec{E}$ is in the negative $x$ - direction, the potential increases in the positive $x$ - direction. Given $x_a = 0.60\ m$ and $x_b=0.90\ m$, point $b$ is at a higher potential.

Step2: Calculate the electric - field magnitude

The relationship between the electric field $E$, the potential difference $\Delta V$, and the displacement $d$ in a uniform electric field is $\Delta V = Ed$. Here, $\Delta V=V_a - V_b=- 240\ V$ (negative because $V_b>V_a$) and $d=x_b - x_a=0.90\ m - 0.60\ m = 0.30\ m$. Using $E=\frac{|\Delta V|}{d}$, we have $E=\frac{240\ V}{0.30\ m}=800\ V/m$.

Step3: Calculate the work done

The work done by the electric field $W$ on a charge $q$ moving through a potential difference $\Delta V$ is $W = q\Delta V$. Here, $q=-0.200\times10^{- 6}\ C$ and $\Delta V=-240\ V$. So $W=(-0.200\times10^{-6}\ C)\times(-240\ V)=4.8\times10^{-5}\ J$.

Answer:

(a) Point $b$ is at the higher potential. (b) $E = 800\ V/m$ (c) $W = 4.8\times10^{-5}\ J$