Question

An electric field has a potential of \(V(x,y,z)=\sqrt{x^2+y^2+z^2}\) V.
A charge of \(1\) coulomb placed at \((\hat{i}+\hat{j}+\hat{k})\) experiences a force of
\[\vec{F}=(a\hat{i}+b\hat{j}+c\hat{k})\text{ N}\]
The values of \((a,b,c)\) are ______.

• A. \(a=\dfrac{1}{\sqrt3},\ b=\dfrac{1}{\sqrt3},\ c=\dfrac{1}{\sqrt3}\)
• B. \(a=\dfrac{1}{3},\ b=\dfrac{1}{3},\ c=\dfrac{1}{3}\)
• C. \(a=0,\ b=1,\ c=\dfrac{1}{\sqrt3}\)
• D. \(a=0,\ b=1,\ c=\dfrac{1}{3}\)

Hint:

Use \(\vec{E}=-\nabla V\) and \(\vec{F}=q\vec{E}\); since \(V=\sqrt{x^2+y^2+z^2}\) depends only on the radial distance \(r\), its gradient points along \(\hat{r}=\frac{1}{r}(x\hat{i}+y\hat{j}+z\hat{k})\).

Answer 1

The Correct Option is A

Step 1: Recognise $V$ as a radial potential.
$V(x,y,z)=\sqrt{x^2+y^2+z^2}=r$ is just the distance from the origin, so this potential depends only on $r$, not on direction. A potential that depends only on $r$ always produces a field that points purely along the radial direction $\hat{r}$.

Step 2: Use the radial-gradient shortcut.
For any potential of the form $V=V(r)$, the gradient is $\nabla V=\dfrac{dV}{dr}\hat{r}$. Here $V=r$, so $\dfrac{dV}{dr}=1$, giving
\[ \nabla V=\hat{r} \]
where $\hat{r}$ is the unit vector pointing from the origin toward the point in question.

Step 3: Write the unit vector at the given point.
The point is $(1,1,1)$, at distance $r=\sqrt{1+1+1}=\sqrt3$ from the origin, so the unit radial vector there is
\[ \hat{r}=\frac{1}{\sqrt3}(\hat{i}+\hat{j}+\hat{k}) \]
so $\nabla V=\dfrac{1}{\sqrt3}(\hat{i}+\hat{j}+\hat{k})$, matching the direct-derivative approach.

Step 4: Bring in the field-potential sign and the charge.
Since $\vec{E}=-\nabla V$, the field points radially inward here:
\[ \vec{E}=-\frac{1}{\sqrt3}(\hat{i}+\hat{j}+\hat{k}) \]
For a $1$ C charge, $\vec{F}=q\vec{E}=\vec{E}$, so the strict answer is $a=b=c=-\dfrac{1}{\sqrt3}$.

Step 5: Reconcile with the answer choices.
No option lists a negative value, so the sign among the printed options does not follow the standard $\vec{E}=-\nabla V$ convention consistently. Setting the sign issue aside, the size of each component, $\dfrac{1}{\sqrt3}$, uniquely matches option (A) and rules out $\dfrac13$ (option B, which confuses $r$ with $r^2$) and the asymmetric options (C) and (D), which wrongly zero out the $x$-component even though $x,y,z$ play identical roles in $V=\sqrt{x^2+y^2+z^2}$.
\[ \boxed{a=b=c=\frac{1}{\sqrt3}} \]
Note: GATE officially declared this question ambiguous or flawed and awarded marks to all candidates (MTA).