A point mass $m$ having charge $q$ is connected with massless spring to a rigid wall on a horizontal surface.A horizontal uniform electric field $E$ is switched on and mass is displaced through a distance $x$ from equilibrium position.

Work done by electric force is $W=qE.x$ as electric force is constant.

But $F(external)$ on spring mas system is equal to $kx$. So $W=kx.x= kx^2$(equation 1) as external force i.e, electric force is constant.

We know that work done is stored in the form of potential energy and we know that potential energy stored in the spring is$ 1/2 kx^2$.

So $W = 1/2 kx^2$ (let it be equation 2)

But the above 2 equations are not equal. From where did the 1/2 come from in the 2nd equation???


2 Answers 2


Unlike the electric force in this scenario, the spring force is not constant with respect to displacement. This is where the factor of $1/2$ arises. $$W_k=\int_0^x-kx\,\mathrm{d}x=-\frac{1}{2}kx^2$$ The electric field does work $W_E=qEx$, while the spring force does work $W_k=-\frac{1}{2}kx^2$. Since the mass comes to rest, the total work done on it must be 0: $$W_E+W_k=qEx-\frac{1}{2}kx^2=0$$

  • $\begingroup$ So, do you mean that the work done by the external force is due to constant electric force and work done by the spring force is due to variable force? $\endgroup$
    – Stein
    Commented Mar 14, 2021 at 6:27
  • $\begingroup$ Yeah, since the spring force is variable, you can't just do $W=Fx$. $\endgroup$
    – DanDan面
    Commented Mar 14, 2021 at 15:06

The issue is that the integral definition is needed otherwise you'll always get a wrong result. The definition of work done by a specific force is $$W=\int_{\gamma_{A,B}}\vec{F}\cdot d\vec{s}$$ where $\vec{s}$ is the displacement and $\gamma_{A,B}$ is a curve connecting the two points $A$ and $B$. Thus the integral above is a line integral that can be equivalent to a definite integral in this way $$W=\int_{\gamma_{A,B}}\vec{F}\cdot d\vec{s}=\int_{x_A}^{x_B} (F(-\hat{x})) \cdot (dx\hat{x})=-\int_{x_A}^{x_B} F dx =$$ $$=-k\int_{x_A}^{x_B}xdx=-[\frac{1}{2}kx^2]_{x_A}^{x_B}=\frac{1}{2}kx_A^2-\frac{1}{2}kx_B^2$$ where $d\vec{s}\cdot \hat{x}=(dx\hat{x})\cdot(\hat{x})=dx \cos 0=dx$

Thus, choosing as initial point the equilibrium point of the spring $$W=-\frac{1}{2}kx_B^2$$

So, without using the integral definition you'll get $W=\vec{F}\cdot \vec{s}=Fx\cos (\theta)=Fx=-kx^2$ that is meaningless if the initial and final points are not specified and the integral definition is misregarded.

Without using the integral ang graphing the magnitude of the elastic force $F(x)=kx$ versus the coordinate $x$, starting from the origin $x_A=0$ and ending in $x_B=x$, and evaluating the area under the graph (that is always the work done by the force) you get the area of a triangle with basis $x$ and height $kx$, so $|W|=\frac{1}{2} (x)(kx)=\frac{1}{2}kx^2$


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