Given electric field

$\vec E = 30x^2\,\hat i$

This field is along the $x$-direction only.

Relation between electric field and potential

We use

$\vec E = -\nabla V$

Along the $x$-axis,

$E_x = -\frac{dV}{dx}$

So,

$\frac{dV}{dx} = -30x^2$

Find potential difference between $O$ and $A$

We need

$V_A - V_O = V(2)-V(0)$

Therefore,

$V_A - V_O = \int_0^2 \frac{dV}{dx}\,dx = \int_0^2 (-30x^2)\,dx$

$V_A - V_O = -30\int_0^2 x^2\,dx$

$= -30\left[\frac{x^3}{3}\right]_0^2$

$= -30\cdot \frac{8}{3}$

$= -80\,\text{V}$

Since $1\,\text{V} = 1\,\text{J/C}$,

$V_A - V_O = -80\,\text{J/C}$

Check options

• A: $120\,\text{J/C}$ ❌
• B: $-120\,\text{J/C}$ ❌
• C: $-80\,\text{J/C}$ ✅
• D: $80\,\text{J/C}$ ❌

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.