(Solved): A region has a uniform electric field of \( 50 \mathrm{~V} / \mathrm{m} \) to the left. Point \( B ...

A region has a uniform electric field of \( 50 \mathrm{~V} / \mathrm{m} \) to the left. Point \( B \) is 3 meter to the left of, and 4 meters above, point A. The voltage at point \( A \) is \( 600 \mathrm{~V} \). What is the voltage at point \( B \) ? Drow it? \( 800 \mathrm{~V} \) \( 450 \mathrm{~V} \) \( 850 \mathrm{~V} \) \( 350 \mathrm{~V} \) \( 750 \mathrm{~V} \) \( 400 \mathrm{~V} \) The electric potential at the midpoint between the two charges due to one of the \( -Q \) is charges is \( -V_{0} \), while the magnitude of the electric field at the same point due to one of the \( +Q \) charges is \( E_{0} \). What is the voltage and \( x \) component of the electric field at the midpoint between the two charges? Take \( V=0 \) at infinity and assume no other source of electric field. \[ \begin{array}{l} V=-2 V_{0} \text { and } E_{x}=2 E_{0} \\ V=-2 V_{\rho} \text { and } E_{x}=-2 E_{\sigma} \\ V=-2 V_{0} \text { and } E_{x}=0 \\ V=0 \text { and } E_{x}=2 E_{0} \\ V=0 \text { and } E_{x}=0 . \\ V=0 \text { and } E_{x}=-2 E_{r} \end{array} \] A thin rod has uniform charge per length 3 w over its length \( H \). The distance between point \( A \) and point \( B \) is \( 3 H \) and the distance between point \( A \) and point \( P \). is \( 2 \mathrm{H}_{1} \) We introduce an integration variable s with \( s=0 \) chosen to be at point B and the \( + \) s direction to be down. The small red segment has length ds and charge dq. We want to find the electric potential at point P. Drowit out whabel the ali the lengths and the integrotion warioble! What is the charge dq in the small segment ds? Which expression below gives the voltage \( d V \) from the small charge dq in the small segment ds? Choose from the choices (A thru fl below: A. \( d V=\frac{K d-}{\left(g^{2}+4 H^{2}\right)} \) B. \( \quad d V=\frac{R i_{l}}{s} \) C. \( \mathrm{dV}=\frac{\mathrm{K} d \mathrm{~d}_{7}}{\sqrt{A^{3}+4 H^{2}}} \) D. \( d V=\frac{\text { Nido }}{\sqrt{(\theta+3 m)^{3}+4 m^{7}}} \) E. \( d i=\frac{K d 0}{\left((++3 H)^{2}+4 H^{2}\right)} \) F. \( d V=\frac{3 K^{2} d y}{\left((++6 H)^{7}+9 H^{2}\right)^{3 / T}} \)