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: Recall how electric field relates to potential.
The electric field is the negative gradient of the electric potential:
\[ \vec{E}=-\nabla V=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right) \] The minus sign is there because the field points from high potential to low potential, in the direction the potential decreases fastest, while the gradient itself points toward the steepest increase.

Step 2: Differentiate the given potential.
With \(V=\sqrt{x^2+y^2+z^2}=r\), where \(r\) is the distance from the origin,
\[ \frac{\partial V}{\partial x}=\frac{x}{\sqrt{x^2+y^2+z^2}}=\frac{x}{r} \] and in the same way
\[ \frac{\partial V}{\partial y}=\frac{y}{r},\qquad \frac{\partial V}{\partial z}=\frac{z}{r} \] So
\[ \nabla V=\frac{1}{r}(x\hat{i}+y\hat{j}+z\hat{k}) \]

Step 3: Evaluate at the given point.
At \((x,y,z)=(1,1,1)\),
\[ r=\sqrt{1^2+1^2+1^2}=\sqrt3 \] so
\[ \nabla V\Big|_{(1,1,1)}=\frac{1}{\sqrt3}(\hat{i}+\hat{j}+\hat{k}) \] and hence
\[ \vec{E}=-\nabla V=-\frac{1}{\sqrt3}(\hat{i}+\hat{j}+\hat{k}) \]

Step 4: Get the force on the charge.
The force on a charge \(q\) in a field \(\vec{E}\) is \(\vec{F}=q\vec{E}\). With \(q=1\) C,
\[ \vec{F}=1\times\left(-\frac{1}{\sqrt3}\right)(\hat{i}+\hat{j}+\hat{k})=-\frac{1}{\sqrt3}\hat{i}-\frac{1}{\sqrt3}\hat{j}-\frac{1}{\sqrt3}\hat{k} \] So, by the strict sign convention \(\vec{E}=-\nabla V\), the components are \(a=b=c=-\dfrac{1}{\sqrt3}\).

Step 5: Compare with the listed options.
Every listed option gives positive values of \(a,b,c\), while the rigorous calculation above gives \(a=b=c=-\dfrac{1}{\sqrt3}\). The magnitude \(\dfrac{1}{\sqrt3}\) matches option (A) exactly, but the sign does not, since option (A) prints \(+\dfrac{1}{\sqrt3}\) for all three components. None of the four printed options is an exact match once the sign in \(\vec{E}=-\nabla V\) is respected.

Step 6: Rule out options (B), (C) and (D) on magnitude and symmetry grounds.

(B) \(a=b=c=\frac13\): This would come from mistakenly dividing by \(r^2=3\) instead of \(r=\sqrt3\); the magnitude is wrong. Incorrect.

(C) and (D): Both set \(a=0\), but by symmetry of \(V=\sqrt{x^2+y^2+z^2}\) in \(x,y,z\), the gradient (and hence the force) must have equal, nonzero components along all three axes at the symmetric point \((1,1,1)\); there is no reason for the \(x\)-component alone to vanish. Incorrect.

Final Answer:
Matching magnitude with the option list, the intended answer is
\[ \boxed{a=b=c=\frac{1}{\sqrt3}} \] which is option (A), the only option whose size agrees with the exact gradient calculation, though its sign does not match the strict \(\vec{E}=-\nabla V\) convention.
Note: GATE officially declared this question ambiguous or flawed and awarded marks to all candidates (MTA).