A electric charge - 10 - -8 -C is placed the point -4m-7m-2m- At the point -1m-3m-2m- the electricpotential will be 18Vfield has no Y-componentfield will be along Z-axispotential will be 1-8V

Question

Toppr Admin · Accepted Answer

-Electric potential-xA0-V-14-x3C0-x3F5-0qr-14-x3C0-x3F5-0q-x221A-4-x2212-1-2-7-x2212-3-2-2-x2212-2-2-10-x2212-84-x3C0-xD7-8-854-xD7-10-x2212-12-xD7-5-18volt-

A electric charge $10^{- 8} C$ is placed at the point $(4 m, 7 m, 2 m)$ . At the point $(1 m, 3 m, 2 m)$ the electric
potential will be $18 V$
field has no $Y$ -component
field will be along $Z$ -axis
potential will be $1.8 V$

Electric potential $, V = \frac{1}{4 π ϵ_{0}} \frac{q}{r} = \frac{1}{4 π ϵ_{0}} \frac{q}{\sqrt{(4 - 1)^{2} + (7 - 3)^{2} + (2 - 2)^{2}}} = \frac{10^{- 8}}{4 π \times 8.854 \times 10^{- 12} \times 5} = 18 v o l t$

A electric charge 10−8C is placed at the point (4m,7m,2m). At the point (1m,3m,2m) the electricpotential will be 18Vfield has no Y-componentfield will be along Z-axispotential will be 1.8V

Electric potential, V=14πϵ0qr=14πϵ0q√(4−1)2+(7−3)2+(2−2)2=10−84π×8.854×10−12×5=18volt

A electric charge $10^{- 8} C$ is placed at the point $(4 m, 7 m, 2 m)$ . At the point $(1 m, 3 m, 2 m)$ the electric
potential will be $18 V$
field has no $Y$ -component
field will be along $Z$ -axis
potential will be $1.8 V$

Electric potential $, V = \frac{1}{4 π ϵ_{0}} \frac{q}{r} = \frac{1}{4 π ϵ_{0}} \frac{q}{\sqrt{(4 - 1)^{2} + (7 - 3)^{2} + (2 - 2)^{2}}} = \frac{10^{- 8}}{4 π \times 8.854 \times 10^{- 12} \times 5} = 18 v o l t$