A particle of mass m$\mathrm{m}$ and charge q$\mathrm{q}$ is fastened to one end A$\mathrm{A}$ of a massless string having equilibrium length $\ell$ , whose other end is fixed at point O$\mathrm{O}$ . The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is

Related Formula

Core Logic

The resultant dipole moment for a system of charges is given by:

Substituting the given values:

Pattern Recognition

For a system of point charges where the net charge is non-zero, the dipole moment depends on the origin. However, taking the standard formula Σ qᵢ rᵢ$\sum q_i r_i$ yields the required mathematical expression directly.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

Related Formula

• Electric field intensity on the axis of a ring of radius R$R$ at axial distance z$z$:

• Condition for maximum electric field on the axis of a ring:

Core Logic

To find where the axial field is maximized, take the first derivative of E$E$ with respect to z$z$ and set it to zero:

We are given:

• Radius of the loop R = a√(2)$R = a\sqrt{2}$

Step 1: Calculate the peak coordinate

Substitute the radius R = a√(2)$R = a\sqrt{2}$ into the maximized coordinate condition:

Thus, the electric field reaches its peak intensity at z = a$z = a$.

Pattern Recognition

Sees: Maximum axial field location of circular charged ring. Trap: Simply picking radius R$R$ directly as the distance value, or omitting the 1/√(2)$1/\sqrt{2}$ scaling factor. Shortcut: The peak of the axial field distribution of any ring always lies at z = R√(2)$z = \frac{R}{\sqrt{2}}$. With R = a√(2)$R = a\sqrt{2}$, z$z$ resolves simply to a√(2)√(2) = a$\frac{a\sqrt{2}}{\sqrt{2}} = a$.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

Related Formula

The electric flux Φ$\Phi$ through an area vector A$\vec{A}$ in a uniform electric field E$\vec{E}$ is:

Core Logic

Given:

• E = 6 i + 5 j + 3 k$\vec{E} = 6\hat{i} + 5\hat{j} + 3\hat{k}$
• A = 30 i$\vec{A} = 30\hat{i}$ (since the surface lies in the YZ$YZ$-plane, its normal area vector points along the X$X$-axis, which is i$\hat{i}$)

Step 1: Compute Dot Product

Using the vector dot product:

Thus, the flux is 180$180$.

Pattern Recognition

Flux through a plane parallel to YZ$YZ$-plane depends only on the x$x$-component of the electric field (Eₓ$E_x$). Therefore, Φ = Eₓ × A = 6 × 30 = 180$\Phi = E_x \times A = 6 \times 30 = 180$ immediately.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

Related Formula

Gauss's Law states:

Core Logic

We have a sphere of radius R = 4a$R = 4a$ centered at (0,0)$(0,0)$.

• Charge 5Q$5Q$ is located at (3a, 0)$(3a, 0)$. Since the distance from origin is 3a lt 4a$3a \lt 4a$, this charge lies inside the sphere.
• Charge -2Q$-2Q$ is located at (-5a, 0)$(-5a, 0)$. Since the distance from origin is 5a gt 4a$5a \gt 4a$, this charge lies outside the sphere.

Step 1: Calculate Enclosed Charge

The net enclosed charge qenclosed$q_{\text{enclosed}}$ within the spherical boundary is:

Step 2: Apply Gauss Law

Using Gauss's Law, the total electric flux is:

Therefore, the electric flux is (5Q)/(ε₀)$\frac{5Q}{\varepsilon_0}$.

Pattern Recognition

Flux depends purely on charges situated inside the closed surface. Charges outside the Gaussian surface contribute absolutely zero net flux because every field line entering must also exit.

Chapter Mix

Class 12 Physics: Electric Charges and Fields