Progressive Force fields practice questions
Use Clevolab for progressive practice in force fields. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats force fields as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.
The current Progressive set covers gravitational and electric fields, field strength, and potential ideas. 142 reviewed questions currently published for this page.
What you can practise
- Field lines and interpretation
- Field strength
- Potential ideas
- Gravitational and electric comparisons
Real sample questions from the current set
These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.
Sample question
In a metal conductor, the drift direction of electrons compared to conventional current is
- AThe sameAnswer option
- BRandomAnswer option
- COppositeCorrect answer
- DPerpendicularAnswer option
Why this answer is right
Conventional current is defined as the flow of positive charge. Since electrons are negative, they drift opposite to the current direction.
Current is $$I=\frac{dq}{dt}$$ in the direction positive charges would move. Electrons carry charge $-e$, so their motion opposite to $I$ corresponds to a positive current. Thus electron drift is opposite the conventional current.
Sample question
Two charges $+Q$ are at $x=\pm a$. What is the electric potential at the origin?
- A$0$Answer option
- B$kQ/a$Answer option
- C$kQ/(2a)$Answer option
- D$2kQ/a$Correct answer
Why this answer is right
Potential is a scalar and adds algebraically: $V=kQ/a+kQ/a=2kQ/a$. Unlike the field, it does not cancel by symmetry.
For a point charge, $$V=k\frac{Q}{r}.$$ Each charge is a distance $a$ from the origin, so each contributes $kQ/a$. Adding the contributions gives $$V=\frac{kQ}{a}+\frac{kQ}{a}=\frac{2kQ}{a}.$$
Sample question
An infinite line charge of density $\lambda$ lies along the $z$-axis in dielectric $\varepsilon$. What is $V(r_2)-V(r_1)$ for radii $r_2>r_1$?
- A$-\dfrac{\lambda}{2\pi\varepsilon}\,\dfrac{1}{r_2-r_1}$Answer option
- B$-\dfrac{\lambda}{2\pi\varepsilon}\,\ln\!\left(\dfrac{r_2}{r_1}\right)$Correct answer
- C$+\dfrac{\lambda}{4\pi\varepsilon}\,\ln\!\left(\dfrac{r_2}{r_1}\right)$Answer option
- D$0$Answer option
Why this answer is right
With $E_r=\dfrac{\lambda}{2\pi\varepsilon r}$, $dV=-E_r\,dr$. Integrating from $r_1$ to $r_2$ yields $V(r_2)-V(r_1)=-\dfrac{\lambda}{2\pi\varepsilon}\ln\!\left(\dfrac{r_2}{r_1}\right)$.
Gauss’s law gives $$E_r(r)=\frac{\lambda}{2\pi\varepsilon r}.$$
How this page fits into Clevolab
Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.