- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
Variation of electrostatic potential along $x$-direction is shown in the graph. The correct statement about electric field is
A
$x$ component at point $B$ is maximum
B
$x $ component at point $A$ is towards positive $x$-axis.
C
$x$ component at point $C$ is along negative $x-$ axis
D
$x $ component at point $C$ is along positive $x$ -axis
Solution
Electric field $E_{x}=-\frac{d V}{d x} .$ So for negative slope, field will be directed along positive $x$. Thus $x$ component at point $A$ is towards negative $x$ $-axis$ and $x$ component at point $C$ is along positive $x$ $-axis$ and at point $B, x$ $-component$ of field will be minimum.
Standard 12
Physics
