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In a uniform electric field the potential at origin is V and V/2 at points (a,0,0) , (0,b,0) and (0,0,c) .then find the potential at point ( a,b,c) . I don't really get how are we supposed to solve such type of questions I always get confused. A hint would be enough

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Since the electric field is uniform, if you move $l$ meters in a direction that makes an angle of $\theta$ with the electric field, the electric potential rises by an amount of $E\times l \times \cos(\theta)$. So we can say that:

$-\dfrac{V}{2}=a\times E\times \sin(\theta_1)=b\times E\times \sin(\theta_2)=c\times E\times \sin(\theta_3)$

So in order to solve this question, I move $a$ meters in the $x$ axis, which will decrease the electric potential by $\dfrac{V}{2}$, then I move $b$ meters in the $y$ axis, which will again decrease the electric potential by $\dfrac{V}{2}$(because the term $b \times E\times \sin(\theta_2)$ doesn't depend on the position of the object that is moving, it only depends on the movement vector). Now I move $c$ meters in the $z$ axis which will also decrease the potential by $\dfrac{V}{2}$. So the final answer would be $-\dfrac{V}{2}$


Update:

The question says that if we move $a$ meters in the $x$ axis, the potential decreases by $\dfrac{V}{2}$. Because at the origin it was $V$ and at $(a, 0, 0)$ it is $\dfrac{V}{2}$. And the important thing is that since $E$ is uniform, the difference in potential doesn't depend on the place you are, it solely depends on the movement vector. For example if I move from $(a, 0, 0)$ to $(2a, 0, 0)$ then the potential decreases by $\dfrac{V}{2}$ so the potential at $(2a, 0, 0)$ is $0$. Now because we want to calculate the potential at $(a, b, c)$ we can move $a$ meters in the $x$ axis, $b$ meters in the $y$ axis and then move $c$ meters in the $z$ axis. In each part, the potential is decreasing by $\dfrac{V}{2}$. And the initiating potential(the potential at the origin) is $V$. So the final potential is:

$V - \dfrac{V}{2} - \dfrac{V}{2} - \dfrac{V}{2} = -\dfrac{V}{2}$

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  • $\begingroup$ How is the final answer -V/2 if it decreases 3 times? And I didn't get how u wrote the potential equations. $\endgroup$ – Lakshya Gupta May 11 '17 at 10:05
  • $\begingroup$ Hi, I have updated the answer please take a look $\endgroup$ – Soroush khoubyarian May 12 '17 at 8:02
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In a uniform electric field, the plane which is perpendicular to the electric intensity vector is the plane where potential is the same. The difference of potential along the intensity vector can be calculated as: $\Delta V=E \times \Delta D$, where $\Delta D$ is distance between two points. In this question, three points fix a plane of same potential, and the electric intensity can be calculated by the difference of potential between the origin and the plane. Then you can calculate every point in the space.

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  • $\begingroup$ How do we calculate at any point in space? $\endgroup$ – Lakshya Gupta May 11 '17 at 8:51
  • $\begingroup$ @LakshyaGupta Making a line from the origin perpendicular to the plane fixed by three points, you will know the direction of electric field. Calculate the distance between origin and plane, then the intensity can be derived by the formula. Then you know both the intensity and direction of the uniform electric field and the field is uniquely fixed. $\endgroup$ – wsjyhaozi May 11 '17 at 8:59
  • $\begingroup$ @LakshyaGupta sorry i made a mistake on the formula, and i have already corrected it. $\endgroup$ – wsjyhaozi May 11 '17 at 9:00

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